Computing marginal distribution












0












$begingroup$


Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.



The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?



I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?



I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...






real-analysis probability integration probability-theory density-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
    $endgroup$
    – JimB
    Jan 11 at 7:03










  • $begingroup$
    Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
    $endgroup$
    – JimB
    Jan 11 at 17:07


















0












$begingroup$


Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.



The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?



I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?



I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...






real-analysis probability integration probability-theory density-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
    $endgroup$
    – JimB
    Jan 11 at 7:03










  • $begingroup$
    Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
    $endgroup$
    – JimB
    Jan 11 at 17:07
















0












0








0





$begingroup$


Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.



The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?



I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?



I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...






real-analysis probability integration probability-theory density-function







share|cite|improve this question









$endgroup$




Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.



The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?



I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?



I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...






real-analysis probability integration probability-theory density-function




real-analysis probability integration probability-theory density-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 3:39









user509037user509037

836




836












  • $begingroup$
    Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
    $endgroup$
    – JimB
    Jan 11 at 7:03










  • $begingroup$
    Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
    $endgroup$
    – JimB
    Jan 11 at 17:07




















  • $begingroup$
    Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
    $endgroup$
    – JimB
    Jan 11 at 7:03










  • $begingroup$
    Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
    $endgroup$
    – JimB
    Jan 11 at 17:07


















$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03




$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03












$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07






$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07












1 Answer
1






active

oldest

votes


















1












$begingroup$

First look at the sample space associated with a non-zero joint density:



Sample space for joint probability



The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.



We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.



case where c0 leq 1 - lambda



$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$



case where c0 gt 1 - lambda



$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$



So



$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$



Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:



$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$



Here is what the marginal pdf for $C$ looks like for various values of $lambda$:



PDF of c for various values of lambda






share|cite|improve this answer











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    $begingroup$

    First look at the sample space associated with a non-zero joint density:



    Sample space for joint probability



    The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.



    We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.



    case where c0 leq 1 - lambda



    $$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$



    case where c0 gt 1 - lambda



    $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
    frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$



    So



    $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
    { &
    begin{array}{cc}
    frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
    frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
    end{array}
    \
    end{array}$$



    Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:



    $$f(c)=begin{array}{cc}
    { &
    begin{array}{cc}
    frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
    -frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
    end{array}
    \
    end{array}$$



    Here is what the marginal pdf for $C$ looks like for various values of $lambda$:



    PDF of c for various values of lambda






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      First look at the sample space associated with a non-zero joint density:



      Sample space for joint probability



      The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.



      We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.



      case where c0 leq 1 - lambda



      $$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$



      case where c0 gt 1 - lambda



      $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
      frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$



      So



      $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
      { &
      begin{array}{cc}
      frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
      frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
      end{array}
      \
      end{array}$$



      Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:



      $$f(c)=begin{array}{cc}
      { &
      begin{array}{cc}
      frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
      -frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
      end{array}
      \
      end{array}$$



      Here is what the marginal pdf for $C$ looks like for various values of $lambda$:



      PDF of c for various values of lambda






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        First look at the sample space associated with a non-zero joint density:



        Sample space for joint probability



        The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.



        We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.



        case where c0 leq 1 - lambda



        $$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$



        case where c0 gt 1 - lambda



        $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
        frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$



        So



        $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
        { &
        begin{array}{cc}
        frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
        frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
        end{array}
        \
        end{array}$$



        Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:



        $$f(c)=begin{array}{cc}
        { &
        begin{array}{cc}
        frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
        -frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
        end{array}
        \
        end{array}$$



        Here is what the marginal pdf for $C$ looks like for various values of $lambda$:



        PDF of c for various values of lambda






        share|cite|improve this answer











        $endgroup$



        First look at the sample space associated with a non-zero joint density:



        Sample space for joint probability



        The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.



        We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.



        case where c0 leq 1 - lambda



        $$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$



        case where c0 gt 1 - lambda



        $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
        frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$



        So



        $$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
        { &
        begin{array}{cc}
        frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
        frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
        end{array}
        \
        end{array}$$



        Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:



        $$f(c)=begin{array}{cc}
        { &
        begin{array}{cc}
        frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
        -frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
        end{array}
        \
        end{array}$$



        Here is what the marginal pdf for $C$ looks like for various values of $lambda$:



        PDF of c for various values of lambda







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 5:53

























        answered Jan 12 at 5:19









        JimBJimB

        61547




        61547






























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