Continuous Random Variable: Distribution Function & Expected Value
$begingroup$
Given
$$ F(x) = begin{cases}
0, & x < 1 \
c(x-1)^2, & 1 leq x < 2 \
1, & x geq 2
end{cases} $$
- Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.
- Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.
Solution
$F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.- 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
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
$endgroup$
add a comment |
$begingroup$
Given
$$ F(x) = begin{cases}
0, & x < 1 \
c(x-1)^2, & 1 leq x < 2 \
1, & x geq 2
end{cases} $$
- Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.
- Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.
Solution
$F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.- 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
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
$endgroup$
add a comment |
$begingroup$
Given
$$ F(x) = begin{cases}
0, & x < 1 \
c(x-1)^2, & 1 leq x < 2 \
1, & x geq 2
end{cases} $$
- Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.
- Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.
Solution
$F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.- 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
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
$endgroup$
Given
$$ F(x) = begin{cases}
0, & x < 1 \
c(x-1)^2, & 1 leq x < 2 \
1, & x geq 2
end{cases} $$
- Find the values of $c$ for which $F(x)$ is the distribution function of a random variable $X$.
- Establish for what value of $c$ the random variable $X$ is continuous, and in this case derive $E(X)$.
Solution
$F(x)$ is the distribution function of a random variable $X$ for $0 leq c leq 1$.- 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
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
edited Dec 30 '18 at 16:21
Sangchul Lee
94.8k12170276
94.8k12170276
asked Dec 30 '18 at 15:58
1Z101Z10
1105
asked Dec 30 '18 at 15:58
1Z101Z10
1105
asked Dec 30 '18 at 15:58
1Z101Z10
1105
1105
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $0 leq c leq 1$.
$hspace{2em}$
$X$ 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.
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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?
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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 $0 leq c leq 1$.
$hspace{2em}$
$X$ 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.
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $0 leq c leq 1$.
$hspace{2em}$
$X$ 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.
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $0 leq c leq 1$.
$hspace{2em}$
$X$ 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.
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
$endgroup$
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 $0 leq c leq 1$.
$hspace{2em}$
$X$ 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.
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
edited Dec 30 '18 at 16:57
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
answered Dec 30 '18 at 16:31
Sangchul LeeSangchul Lee
94.8k12170276
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
$begingroup$
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?
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
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?
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
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?
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
$endgroup$
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?
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
answered Dec 30 '18 at 16:17
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
