Continuous Random Variable: Distribution Function & Expected Value












0












$begingroup$


Given



$$ F(x) = begin{cases}
0, & x < 1 \
c(x-1)^2, & 1 leq x < 2 \
1, & x geq 2
end{cases} $$




  1. Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.

  2. Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.


Solution





  1. $F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.

  2. For $c=1$ the random variable $X$ is continuous and in this case it results $f(x) = 2(x-1)$ for $1 leq x < 2$ so it results in
    $$ E(X) = int_{1}^{2} 2x(x-1) , mathrm{d}x = 2 left[ frac{x^3}{3} - frac{x^2}{2}right]_{1}^{2} = frac{5}{3} $$


What is the proper way to solve this exercise?
I tried by using



$$ int_{-infty}^{infty} f(x) , mathrm{d}x = 1, $$



but it leads me to a single value of $c=3$, which is not the case of the solution. Is there any formula I'm missing?






probability-distributions random-variables expected-value







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Given



    $$ F(x) = begin{cases}
    0, & x < 1 \
    c(x-1)^2, & 1 leq x < 2 \
    1, & x geq 2
    end{cases} $$




    1. Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.

    2. Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.


    Solution





    1. $F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.

    2. For $c=1$ the random variable $X$ is continuous and in this case it results $f(x) = 2(x-1)$ for $1 leq x < 2$ so it results in
      $$ E(X) = int_{1}^{2} 2x(x-1) , mathrm{d}x = 2 left[ frac{x^3}{3} - frac{x^2}{2}right]_{1}^{2} = frac{5}{3} $$


    What is the proper way to solve this exercise?
    I tried by using



    $$ int_{-infty}^{infty} f(x) , mathrm{d}x = 1, $$



    but it leads me to a single value of $c=3$, which is not the case of the solution. Is there any formula I'm missing?






    probability-distributions random-variables expected-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Given



      $$ F(x) = begin{cases}
      0, & x < 1 \
      c(x-1)^2, & 1 leq x < 2 \
      1, & x geq 2
      end{cases} $$




      1. Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.

      2. Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.


      Solution





      1. $F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.

      2. For $c=1$ the random variable $X$ is continuous and in this case it results $f(x) = 2(x-1)$ for $1 leq x < 2$ so it results in
        $$ E(X) = int_{1}^{2} 2x(x-1) , mathrm{d}x = 2 left[ frac{x^3}{3} - frac{x^2}{2}right]_{1}^{2} = frac{5}{3} $$


      What is the proper way to solve this exercise?
      I tried by using



      $$ int_{-infty}^{infty} f(x) , mathrm{d}x = 1, $$



      but it leads me to a single value of $c=3$, which is not the case of the solution. Is there any formula I'm missing?






      probability-distributions random-variables expected-value







      share|cite|improve this question











      $endgroup$




      Given



      $$ F(x) = begin{cases}
      0, & x < 1 \
      c(x-1)^2, & 1 leq x < 2 \
      1, & x geq 2
      end{cases} $$




      1. Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.

      2. Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.


      Solution





      1. $F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.

      2. For $c=1$ the random variable $X$ is continuous and in this case it results $f(x) = 2(x-1)$ for $1 leq x < 2$ so it results in
        $$ E(X) = int_{1}^{2} 2x(x-1) , mathrm{d}x = 2 left[ frac{x^3}{3} - frac{x^2}{2}right]_{1}^{2} = frac{5}{3} $$


      What is the proper way to solve this exercise?
      I tried by using



      $$ int_{-infty}^{infty} f(x) , mathrm{d}x = 1, $$



      but it leads me to a single value of $c=3$, which is not the case of the solution. Is there any formula I'm missing?






      probability-distributions random-variables expected-value




      probability-distributions random-variables expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 16:21









      Sangchul Lee

      94.8k12170276




      94.8k12170276










      asked Dec 30 '18 at 15:58









      1Z101Z10

      1105




      1105






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$



          1. Recall that $F$ is a distribution function (a.k.a. c.d.f.) if and only if the following properties hold:






            • $F$ is non-decreasing,


            • $F(x) in [0, 1]$ for all $x in mathbb{R}$,


            • $F$ is right-continuous.




            The $F$ defined as in the problem is always right-continuous, so the last property is not our concern. For the rest, a moment of thought tells that we must have $0 leq lim_{xto 2^-} F(x) leq 1$, which is equivalent to $0 leq c leq 1$.



            $hspace{2em}$ Comparison for different values of c




          2. $X$ is continuous if and only if $F$ is absolutely continuous, which means that



            $$F(x) = int_{-infty}^{x} f(t) , mathrm{d}t$$



            for some non-negative function $f$ and for all $x in mathbb{R}$. A necessary condition is that $F$ itself is continuous. (And practically, this is also a sufficient condition at the level of preliminary probability course, as one never deals with singular distribution at this level.)



            Now, we see that $c = 1$ is the only choice for which $F$ is continuous. And with this choice, we easily identify the p.d.f. $f$ as



            $$ f(x) = begin{cases} 2(x-1), & 1 < x < 2 \ 0, & text{otherwise} end{cases}. $$



            Plugging this to the formula $E(X) = int_{-infty}^{infty} xf(x) , mathrm{d}x$ gives the desired answer.








          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            According to Wikipedia it is enough if $F$ is continuous.
            $endgroup$
            – drhab
            Dec 30 '18 at 16:37










          • $begingroup$
            @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:44












          • $begingroup$
            Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
            $endgroup$
            – 1Z10
            Dec 30 '18 at 16:46






          • 1




            $begingroup$
            @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:48








          • 1




            $begingroup$
            @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:58



















          1












          $begingroup$

          By solving this the PDF should be left out.



          The question is about a CDF which is a function with special properties.



          In 1) it is asked: for what $c$ does this function indeed have these properties?



          In 2) it must be noted that a random variable is continuous iff its CDF is continuous.



          So actually it is asked: for what $c$ found in 1) is the function continuous?






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056958%2fcontinuous-random-variable-distribution-function-expected-value%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$



            1. Recall that $F$ is a distribution function (a.k.a. c.d.f.) if and only if the following properties hold:






              • $F$ is non-decreasing,


              • $F(x) in [0, 1]$ for all $x in mathbb{R}$,


              • $F$ is right-continuous.




              The $F$ defined as in the problem is always right-continuous, so the last property is not our concern. For the rest, a moment of thought tells that we must have $0 leq lim_{xto 2^-} F(x) leq 1$, which is equivalent to $0 leq c leq 1$.



              $hspace{2em}$ Comparison for different values of c




            2. $X$ is continuous if and only if $F$ is absolutely continuous, which means that



              $$F(x) = int_{-infty}^{x} f(t) , mathrm{d}t$$



              for some non-negative function $f$ and for all $x in mathbb{R}$. A necessary condition is that $F$ itself is continuous. (And practically, this is also a sufficient condition at the level of preliminary probability course, as one never deals with singular distribution at this level.)



              Now, we see that $c = 1$ is the only choice for which $F$ is continuous. And with this choice, we easily identify the p.d.f. $f$ as



              $$ f(x) = begin{cases} 2(x-1), & 1 < x < 2 \ 0, & text{otherwise} end{cases}. $$



              Plugging this to the formula $E(X) = int_{-infty}^{infty} xf(x) , mathrm{d}x$ gives the desired answer.








            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              According to Wikipedia it is enough if $F$ is continuous.
              $endgroup$
              – drhab
              Dec 30 '18 at 16:37










            • $begingroup$
              @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:44












            • $begingroup$
              Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
              $endgroup$
              – 1Z10
              Dec 30 '18 at 16:46






            • 1




              $begingroup$
              @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:48








            • 1




              $begingroup$
              @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:58
















            1












            $begingroup$



            1. Recall that $F$ is a distribution function (a.k.a. c.d.f.) if and only if the following properties hold:






              • $F$ is non-decreasing,


              • $F(x) in [0, 1]$ for all $x in mathbb{R}$,


              • $F$ is right-continuous.




              The $F$ defined as in the problem is always right-continuous, so the last property is not our concern. For the rest, a moment of thought tells that we must have $0 leq lim_{xto 2^-} F(x) leq 1$, which is equivalent to $0 leq c leq 1$.



              $hspace{2em}$ Comparison for different values of c




            2. $X$ is continuous if and only if $F$ is absolutely continuous, which means that



              $$F(x) = int_{-infty}^{x} f(t) , mathrm{d}t$$



              for some non-negative function $f$ and for all $x in mathbb{R}$. A necessary condition is that $F$ itself is continuous. (And practically, this is also a sufficient condition at the level of preliminary probability course, as one never deals with singular distribution at this level.)



              Now, we see that $c = 1$ is the only choice for which $F$ is continuous. And with this choice, we easily identify the p.d.f. $f$ as



              $$ f(x) = begin{cases} 2(x-1), & 1 < x < 2 \ 0, & text{otherwise} end{cases}. $$



              Plugging this to the formula $E(X) = int_{-infty}^{infty} xf(x) , mathrm{d}x$ gives the desired answer.








            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              According to Wikipedia it is enough if $F$ is continuous.
              $endgroup$
              – drhab
              Dec 30 '18 at 16:37










            • $begingroup$
              @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:44












            • $begingroup$
              Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
              $endgroup$
              – 1Z10
              Dec 30 '18 at 16:46






            • 1




              $begingroup$
              @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:48








            • 1




              $begingroup$
              @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:58














            1












            1








            1





            $begingroup$



            1. Recall that $F$ is a distribution function (a.k.a. c.d.f.) if and only if the following properties hold:






              • $F$ is non-decreasing,


              • $F(x) in [0, 1]$ for all $x in mathbb{R}$,


              • $F$ is right-continuous.




              The $F$ defined as in the problem is always right-continuous, so the last property is not our concern. For the rest, a moment of thought tells that we must have $0 leq lim_{xto 2^-} F(x) leq 1$, which is equivalent to $0 leq c leq 1$.



              $hspace{2em}$ Comparison for different values of c




            2. $X$ is continuous if and only if $F$ is absolutely continuous, which means that



              $$F(x) = int_{-infty}^{x} f(t) , mathrm{d}t$$



              for some non-negative function $f$ and for all $x in mathbb{R}$. A necessary condition is that $F$ itself is continuous. (And practically, this is also a sufficient condition at the level of preliminary probability course, as one never deals with singular distribution at this level.)



              Now, we see that $c = 1$ is the only choice for which $F$ is continuous. And with this choice, we easily identify the p.d.f. $f$ as



              $$ f(x) = begin{cases} 2(x-1), & 1 < x < 2 \ 0, & text{otherwise} end{cases}. $$



              Plugging this to the formula $E(X) = int_{-infty}^{infty} xf(x) , mathrm{d}x$ gives the desired answer.








            share|cite|improve this answer











            $endgroup$





            1. Recall that $F$ is a distribution function (a.k.a. c.d.f.) if and only if the following properties hold:






              • $F$ is non-decreasing,


              • $F(x) in [0, 1]$ for all $x in mathbb{R}$,


              • $F$ is right-continuous.




              The $F$ defined as in the problem is always right-continuous, so the last property is not our concern. For the rest, a moment of thought tells that we must have $0 leq lim_{xto 2^-} F(x) leq 1$, which is equivalent to $0 leq c leq 1$.



              $hspace{2em}$ Comparison for different values of c




            2. $X$ is continuous if and only if $F$ is absolutely continuous, which means that



              $$F(x) = int_{-infty}^{x} f(t) , mathrm{d}t$$



              for some non-negative function $f$ and for all $x in mathbb{R}$. A necessary condition is that $F$ itself is continuous. (And practically, this is also a sufficient condition at the level of preliminary probability course, as one never deals with singular distribution at this level.)



              Now, we see that $c = 1$ is the only choice for which $F$ is continuous. And with this choice, we easily identify the p.d.f. $f$ as



              $$ f(x) = begin{cases} 2(x-1), & 1 < x < 2 \ 0, & text{otherwise} end{cases}. $$



              Plugging this to the formula $E(X) = int_{-infty}^{infty} xf(x) , mathrm{d}x$ gives the desired answer.









            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 30 '18 at 16:57

























            answered Dec 30 '18 at 16:31









            Sangchul LeeSangchul Lee

            94.8k12170276




            94.8k12170276












            • $begingroup$
              According to Wikipedia it is enough if $F$ is continuous.
              $endgroup$
              – drhab
              Dec 30 '18 at 16:37










            • $begingroup$
              @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:44












            • $begingroup$
              Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
              $endgroup$
              – 1Z10
              Dec 30 '18 at 16:46






            • 1




              $begingroup$
              @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:48








            • 1




              $begingroup$
              @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:58


















            • $begingroup$
              According to Wikipedia it is enough if $F$ is continuous.
              $endgroup$
              – drhab
              Dec 30 '18 at 16:37










            • $begingroup$
              @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:44












            • $begingroup$
              Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
              $endgroup$
              – 1Z10
              Dec 30 '18 at 16:46






            • 1




              $begingroup$
              @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:48








            • 1




              $begingroup$
              @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
              $endgroup$
              – Sangchul Lee
              Dec 30 '18 at 16:58
















            $begingroup$
            According to Wikipedia it is enough if $F$ is continuous.
            $endgroup$
            – drhab
            Dec 30 '18 at 16:37




            $begingroup$
            According to Wikipedia it is enough if $F$ is continuous.
            $endgroup$
            – drhab
            Dec 30 '18 at 16:37












            $begingroup$
            @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:44






            $begingroup$
            @drhab, Perhaps there are different conventions for what it means to be a continuous distribution. But in most literature that I have checked, the definition of continuous distribution does not encompass singular distributions. (See this MIT open course, for instance.)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:44














            $begingroup$
            Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
            $endgroup$
            – 1Z10
            Dec 30 '18 at 16:46




            $begingroup$
            Could you please explain further how do you conclude that c=1 is the only choice for which F is continuous?
            $endgroup$
            – 1Z10
            Dec 30 '18 at 16:46




            1




            1




            $begingroup$
            @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:48






            $begingroup$
            @1Z10, Plotting the graph of $F$ will be convincing enough. But if one needs a solid proof, notice that $x = 2$ is the only possible discontinuity of $F$. Since $F(2) = 1= lim_{xto 2^+}F(x)$ and $lim_{xto2^-}F(x) = c$, it suffices to show that they match. So we obtain the necessary condition $c = 1$.
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:48






            1




            1




            $begingroup$
            @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:58




            $begingroup$
            @1Z10, Glad it helped. I also added plots of $F$ for different values of $c$. :)
            $endgroup$
            – Sangchul Lee
            Dec 30 '18 at 16:58











            1












            $begingroup$

            By solving this the PDF should be left out.



            The question is about a CDF which is a function with special properties.



            In 1) it is asked: for what $c$ does this function indeed have these properties?



            In 2) it must be noted that a random variable is continuous iff its CDF is continuous.



            So actually it is asked: for what $c$ found in 1) is the function continuous?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              By solving this the PDF should be left out.



              The question is about a CDF which is a function with special properties.



              In 1) it is asked: for what $c$ does this function indeed have these properties?



              In 2) it must be noted that a random variable is continuous iff its CDF is continuous.



              So actually it is asked: for what $c$ found in 1) is the function continuous?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                By solving this the PDF should be left out.



                The question is about a CDF which is a function with special properties.



                In 1) it is asked: for what $c$ does this function indeed have these properties?



                In 2) it must be noted that a random variable is continuous iff its CDF is continuous.



                So actually it is asked: for what $c$ found in 1) is the function continuous?






                share|cite|improve this answer









                $endgroup$



                By solving this the PDF should be left out.



                The question is about a CDF which is a function with special properties.



                In 1) it is asked: for what $c$ does this function indeed have these properties?



                In 2) it must be noted that a random variable is continuous iff its CDF is continuous.



                So actually it is asked: for what $c$ found in 1) is the function continuous?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '18 at 16:17









                drhabdrhab

                102k545136




                102k545136






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056958%2fcontinuous-random-variable-distribution-function-expected-value%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more