Computing conditional expenctation of independent uniform rv











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Suppose $X$ and $Y$ are two independent uniform on [0,1] . Compute



$$ E[X^2 mid X+Y = a ] $$



where $a in mathbb{R}$ but $0<a<2$




Try.



First, we can find density of $Z=X^2$. We have



$$ P(X^2 leq z ) = P( - sqrt{z} leq X leq sqrt{z}) = int_{- sqrt{z}}^{sqrt{z}} dx =2 sqrt{z}$$



Therefore,



$$ f_Z(z) = frac{1}{sqrt{z}}$$



Now, here is where the trouble starts since I always get confused when computing the conditional expectation. Do they mean that I need to compute



$$ int int_{sqrt{z}+y=a} z frac{1}{sqrt{z} } dy dz$$



??






probability







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  • 1




    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    – leonbloy
    2 hours ago















up vote
2
down vote

favorite













Suppose $X$ and $Y$ are two independent uniform on [0,1] . Compute



$$ E[X^2 mid X+Y = a ] $$



where $a in mathbb{R}$ but $0<a<2$




Try.



First, we can find density of $Z=X^2$. We have



$$ P(X^2 leq z ) = P( - sqrt{z} leq X leq sqrt{z}) = int_{- sqrt{z}}^{sqrt{z}} dx =2 sqrt{z}$$



Therefore,



$$ f_Z(z) = frac{1}{sqrt{z}}$$



Now, here is where the trouble starts since I always get confused when computing the conditional expectation. Do they mean that I need to compute



$$ int int_{sqrt{z}+y=a} z frac{1}{sqrt{z} } dy dz$$



??






probability







share|cite|improve this question


















  • 1




    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    – leonbloy
    2 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Suppose $X$ and $Y$ are two independent uniform on [0,1] . Compute



$$ E[X^2 mid X+Y = a ] $$



where $a in mathbb{R}$ but $0<a<2$




Try.



First, we can find density of $Z=X^2$. We have



$$ P(X^2 leq z ) = P( - sqrt{z} leq X leq sqrt{z}) = int_{- sqrt{z}}^{sqrt{z}} dx =2 sqrt{z}$$



Therefore,



$$ f_Z(z) = frac{1}{sqrt{z}}$$



Now, here is where the trouble starts since I always get confused when computing the conditional expectation. Do they mean that I need to compute



$$ int int_{sqrt{z}+y=a} z frac{1}{sqrt{z} } dy dz$$



??






probability







share|cite|improve this question














Suppose $X$ and $Y$ are two independent uniform on [0,1] . Compute



$$ E[X^2 mid X+Y = a ] $$



where $a in mathbb{R}$ but $0<a<2$




Try.



First, we can find density of $Z=X^2$. We have



$$ P(X^2 leq z ) = P( - sqrt{z} leq X leq sqrt{z}) = int_{- sqrt{z}}^{sqrt{z}} dx =2 sqrt{z}$$



Therefore,



$$ f_Z(z) = frac{1}{sqrt{z}}$$



Now, here is where the trouble starts since I always get confused when computing the conditional expectation. Do they mean that I need to compute



$$ int int_{sqrt{z}+y=a} z frac{1}{sqrt{z} } dy dz$$



??






probability




probability






share|cite|improve this question













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asked 4 hours ago









Jimmy Sabater

1,846218




1,846218








  • 1




    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    – leonbloy
    2 hours ago














  • 1




    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    – leonbloy
    2 hours ago








1




1




@gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
– leonbloy
2 hours ago




@gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
– leonbloy
2 hours ago










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Find joint density function of $U=X$ and $V=X+Y$ via transformation rule. Then we have
$$
f_{U,V}(u,v) = 1_{{0<u<1,; 0<v-u<1}}.
$$ We can compute conditional pdf $f_{U|V}(u|v)$ as follows.
$$
f_{U|V}(u|v)=frac{f_{U,V}(u,v)}{f_V(v)} = begin{cases}frac{1}{v}1_{{0<u<1,; 0<v-u<1}} text{ for }vin (0,1)\frac{1}{2-v}1_{{0<u<1,; 0<v-u<1}}text{ for }vin (1,2)
end{cases}$$What is left is to actually calculate $E[U^2|V=v]$ as follows.
$$
E[U^2|V=v] = int u^2f_{U|V}(u|v)du = frac{1}{v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{v}int_0^v u^2du = frac{v^2}{3}
$$ for $vin (0,1)$ and
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$






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    Find joint density function of $U=X$ and $V=X+Y$ via transformation rule. Then we have
    $$
    f_{U,V}(u,v) = 1_{{0<u<1,; 0<v-u<1}}.
    $$     We can compute conditional pdf $f_{U|V}(u|v)$ as follows.
    $$
    f_{U|V}(u|v)=frac{f_{U,V}(u,v)}{f_V(v)} = begin{cases}frac{1}{v}1_{{0<u<1,; 0<v-u<1}} text{ for }vin (0,1)\frac{1}{2-v}1_{{0<u<1,; 0<v-u<1}}text{ for }vin (1,2)
    end{cases}$$    What is left is to actually calculate $E[U^2|V=v]$ as follows.
    $$
    E[U^2|V=v] = int u^2f_{U|V}(u|v)du = frac{1}{v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{v}int_0^v u^2du = frac{v^2}{3}
    $$     for $vin (0,1)$ and
    $$
    frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
    $$






    share|cite|improve this answer

























      up vote
      3
      down vote













      Find joint density function of $U=X$ and $V=X+Y$ via transformation rule. Then we have
      $$
      f_{U,V}(u,v) = 1_{{0<u<1,; 0<v-u<1}}.
      $$       We can compute conditional pdf $f_{U|V}(u|v)$ as follows.
      $$
      f_{U|V}(u|v)=frac{f_{U,V}(u,v)}{f_V(v)} = begin{cases}frac{1}{v}1_{{0<u<1,; 0<v-u<1}} text{ for }vin (0,1)\frac{1}{2-v}1_{{0<u<1,; 0<v-u<1}}text{ for }vin (1,2)
      end{cases}$$      What is left is to actually calculate $E[U^2|V=v]$ as follows.
      $$
      E[U^2|V=v] = int u^2f_{U|V}(u|v)du = frac{1}{v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{v}int_0^v u^2du = frac{v^2}{3}
      $$       for $vin (0,1)$ and
      $$
      frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
      $$






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        Find joint density function of $U=X$ and $V=X+Y$ via transformation rule. Then we have
        $$
        f_{U,V}(u,v) = 1_{{0<u<1,; 0<v-u<1}}.
        $$         We can compute conditional pdf $f_{U|V}(u|v)$ as follows.
        $$
        f_{U|V}(u|v)=frac{f_{U,V}(u,v)}{f_V(v)} = begin{cases}frac{1}{v}1_{{0<u<1,; 0<v-u<1}} text{ for }vin (0,1)\frac{1}{2-v}1_{{0<u<1,; 0<v-u<1}}text{ for }vin (1,2)
        end{cases}$$        What is left is to actually calculate $E[U^2|V=v]$ as follows.
        $$
        E[U^2|V=v] = int u^2f_{U|V}(u|v)du = frac{1}{v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{v}int_0^v u^2du = frac{v^2}{3}
        $$         for $vin (0,1)$ and
        $$
        frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
        $$






        share|cite|improve this answer












        Find joint density function of $U=X$ and $V=X+Y$ via transformation rule. Then we have
        $$
        f_{U,V}(u,v) = 1_{{0<u<1,; 0<v-u<1}}.
        $$         We can compute conditional pdf $f_{U|V}(u|v)$ as follows.
        $$
        f_{U|V}(u|v)=frac{f_{U,V}(u,v)}{f_V(v)} = begin{cases}frac{1}{v}1_{{0<u<1,; 0<v-u<1}} text{ for }vin (0,1)\frac{1}{2-v}1_{{0<u<1,; 0<v-u<1}}text{ for }vin (1,2)
        end{cases}$$        What is left is to actually calculate $E[U^2|V=v]$ as follows.
        $$
        E[U^2|V=v] = int u^2f_{U|V}(u|v)du = frac{1}{v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{v}int_0^v u^2du = frac{v^2}{3}
        $$         for $vin (0,1)$ and
        $$
        frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
        $$







        share|cite|improve this answer












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        share|cite|improve this answer










        answered 2 hours ago









        Song

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