Finding the Conditional Distribution of a Random Variable Given Another Random Variable of a Different...











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This is the problem i'm currently tackling from my textbook:




Let $P$ have a uniform distribution on $[0,1]$, and, conditional on $P=p$, let $X$ have a Bernoulli distribution with parameter $p$. Find the conditional distribution of $P$ given $X$.




However, while I would be grateful for the solution to this question as a means to check when I (hopefully) finish the question. I'm more interested in the general approach to questions like these. Something of the sort: Let $X$ have a [distribution], conditional on $X=x$, let $Y$ have a [different distribution]. Find the conditional distribution of $X$ given $Y$.



Maybe i'm being too vague, but is there a general/systematic approach to these sort of questions? Like, I know that we can have it so that:



$f_{X|Y}(x|y)=frac{f_{X,Y}(x,y)}{f_Y(y)}$



but then what would you do afterwards and how would you get a solution from that? Would you just sub in the appropriate density into $f_Y(y)$, e.g. if $Y$ had an exponential distribution, would $f_Y(y)=λe^{-λy}$? And what would you do with the joint distribution?



I hope what i'm saying makes sense...please let me know if there's anything I should clarify.










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  • This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
    – Michael
    yesterday

















up vote
3
down vote

favorite
1












This is the problem i'm currently tackling from my textbook:




Let $P$ have a uniform distribution on $[0,1]$, and, conditional on $P=p$, let $X$ have a Bernoulli distribution with parameter $p$. Find the conditional distribution of $P$ given $X$.




However, while I would be grateful for the solution to this question as a means to check when I (hopefully) finish the question. I'm more interested in the general approach to questions like these. Something of the sort: Let $X$ have a [distribution], conditional on $X=x$, let $Y$ have a [different distribution]. Find the conditional distribution of $X$ given $Y$.



Maybe i'm being too vague, but is there a general/systematic approach to these sort of questions? Like, I know that we can have it so that:



$f_{X|Y}(x|y)=frac{f_{X,Y}(x,y)}{f_Y(y)}$



but then what would you do afterwards and how would you get a solution from that? Would you just sub in the appropriate density into $f_Y(y)$, e.g. if $Y$ had an exponential distribution, would $f_Y(y)=λe^{-λy}$? And what would you do with the joint distribution?



I hope what i'm saying makes sense...please let me know if there's anything I should clarify.










share|cite|improve this question









New contributor




BoilingKettle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
    – Michael
    yesterday















up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





This is the problem i'm currently tackling from my textbook:




Let $P$ have a uniform distribution on $[0,1]$, and, conditional on $P=p$, let $X$ have a Bernoulli distribution with parameter $p$. Find the conditional distribution of $P$ given $X$.




However, while I would be grateful for the solution to this question as a means to check when I (hopefully) finish the question. I'm more interested in the general approach to questions like these. Something of the sort: Let $X$ have a [distribution], conditional on $X=x$, let $Y$ have a [different distribution]. Find the conditional distribution of $X$ given $Y$.



Maybe i'm being too vague, but is there a general/systematic approach to these sort of questions? Like, I know that we can have it so that:



$f_{X|Y}(x|y)=frac{f_{X,Y}(x,y)}{f_Y(y)}$



but then what would you do afterwards and how would you get a solution from that? Would you just sub in the appropriate density into $f_Y(y)$, e.g. if $Y$ had an exponential distribution, would $f_Y(y)=λe^{-λy}$? And what would you do with the joint distribution?



I hope what i'm saying makes sense...please let me know if there's anything I should clarify.










share|cite|improve this question









New contributor




BoilingKettle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











This is the problem i'm currently tackling from my textbook:




Let $P$ have a uniform distribution on $[0,1]$, and, conditional on $P=p$, let $X$ have a Bernoulli distribution with parameter $p$. Find the conditional distribution of $P$ given $X$.




However, while I would be grateful for the solution to this question as a means to check when I (hopefully) finish the question. I'm more interested in the general approach to questions like these. Something of the sort: Let $X$ have a [distribution], conditional on $X=x$, let $Y$ have a [different distribution]. Find the conditional distribution of $X$ given $Y$.



Maybe i'm being too vague, but is there a general/systematic approach to these sort of questions? Like, I know that we can have it so that:



$f_{X|Y}(x|y)=frac{f_{X,Y}(x,y)}{f_Y(y)}$



but then what would you do afterwards and how would you get a solution from that? Would you just sub in the appropriate density into $f_Y(y)$, e.g. if $Y$ had an exponential distribution, would $f_Y(y)=λe^{-λy}$? And what would you do with the joint distribution?



I hope what i'm saying makes sense...please let me know if there's anything I should clarify.







probability probability-distributions conditional-probability






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edited yesterday









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asked yesterday









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BoilingKettle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
    – Michael
    yesterday




















  • This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
    – Michael
    yesterday


















This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
– Michael
yesterday






This is a mix of continuous and discrete. $$f_{P|X=x}(p|X=x) = frac{P[X=x|P=p]f_P(p)}{P[X=x]}$$ and $P[X=x] = int_{-infty}^{infty} P[X=x|P=p]f_P(p)dp$
– Michael
yesterday












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  1. We have
    $$ Xsim f_X$$
    and
    $$ Y|X=x sim f_{Y|X}(cdot|x).$$
    We want
    $$ X|Y=ysim f_{X|Y}(cdot|y)=?$$


  2. We know
    $$ f_{X|Y=y}(cdot|y)=frac{f_{X,Y}(cdot,y)}{f_Y(y)}.$$
    It is enough to determine the joint distribution.


  3. Of (1) we have
    $$ f_{X,Y}(x,cdot)= f_{Y|X=x}(cdot)f_X(x).$$







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    1. We have
      $$ Xsim f_X$$
      and
      $$ Y|X=x sim f_{Y|X}(cdot|x).$$
      We want
      $$ X|Y=ysim f_{X|Y}(cdot|y)=?$$


    2. We know
      $$ f_{X|Y=y}(cdot|y)=frac{f_{X,Y}(cdot,y)}{f_Y(y)}.$$
      It is enough to determine the joint distribution.


    3. Of (1) we have
      $$ f_{X,Y}(x,cdot)= f_{Y|X=x}(cdot)f_X(x).$$







    share|cite|improve this answer

























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      0
      down vote














      1. We have
        $$ Xsim f_X$$
        and
        $$ Y|X=x sim f_{Y|X}(cdot|x).$$
        We want
        $$ X|Y=ysim f_{X|Y}(cdot|y)=?$$


      2. We know
        $$ f_{X|Y=y}(cdot|y)=frac{f_{X,Y}(cdot,y)}{f_Y(y)}.$$
        It is enough to determine the joint distribution.


      3. Of (1) we have
        $$ f_{X,Y}(x,cdot)= f_{Y|X=x}(cdot)f_X(x).$$







      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote










        1. We have
          $$ Xsim f_X$$
          and
          $$ Y|X=x sim f_{Y|X}(cdot|x).$$
          We want
          $$ X|Y=ysim f_{X|Y}(cdot|y)=?$$


        2. We know
          $$ f_{X|Y=y}(cdot|y)=frac{f_{X,Y}(cdot,y)}{f_Y(y)}.$$
          It is enough to determine the joint distribution.


        3. Of (1) we have
          $$ f_{X,Y}(x,cdot)= f_{Y|X=x}(cdot)f_X(x).$$







        share|cite|improve this answer













        1. We have
          $$ Xsim f_X$$
          and
          $$ Y|X=x sim f_{Y|X}(cdot|x).$$
          We want
          $$ X|Y=ysim f_{X|Y}(cdot|y)=?$$


        2. We know
          $$ f_{X|Y=y}(cdot|y)=frac{f_{X,Y}(cdot,y)}{f_Y(y)}.$$
          It is enough to determine the joint distribution.


        3. Of (1) we have
          $$ f_{X,Y}(x,cdot)= f_{Y|X=x}(cdot)f_X(x).$$








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        share|cite|improve this answer










        answered 1 hour ago









        Daniel Camarena Perez

        39718




        39718






















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