Computing conditional expectation of independent uniform rv












3












$begingroup$



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



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



where $ain (0,2)$.




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-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    $endgroup$
    – leonbloy
    Dec 16 '18 at 3:43
















3












$begingroup$



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



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



where $ain (0,2)$.




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-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    $endgroup$
    – leonbloy
    Dec 16 '18 at 3:43














3












3








3


1



$begingroup$



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



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



where $ain (0,2)$.




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-distributions







share|cite|improve this question











$endgroup$





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



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



where $ain (0,2)$.




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-distributions




probability probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 19:21









Ethan Bolker

45.4k553120




45.4k553120










asked Dec 16 '18 at 2:19









Jimmy SabaterJimmy Sabater

2,560325




2,560325








  • 1




    $begingroup$
    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    $endgroup$
    – leonbloy
    Dec 16 '18 at 3:43














  • 1




    $begingroup$
    @gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
    $endgroup$
    – leonbloy
    Dec 16 '18 at 3:43








1




1




$begingroup$
@gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
$endgroup$
– leonbloy
Dec 16 '18 at 3:43




$begingroup$
@gd1035 That's wrong. Actually it should be $E([a-Y]^2 | X+Y=a)$ .. which of course is not easier than the original :-)
$endgroup$
– leonbloy
Dec 16 '18 at 3:43










1 Answer
1






active

oldest

votes


















6












$begingroup$

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









$endgroup$













  • $begingroup$
    It's very nice!! Thank you!!
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:58










  • $begingroup$
    but, question, why do you condition on $V=v$? Isnt it a constant $a$?
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:59










  • $begingroup$
    Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
    $endgroup$
    – Song
    Dec 16 '18 at 19:09











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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









$endgroup$













  • $begingroup$
    It's very nice!! Thank you!!
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:58










  • $begingroup$
    but, question, why do you condition on $V=v$? Isnt it a constant $a$?
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:59










  • $begingroup$
    Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
    $endgroup$
    – Song
    Dec 16 '18 at 19:09
















6












$begingroup$

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









$endgroup$













  • $begingroup$
    It's very nice!! Thank you!!
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:58










  • $begingroup$
    but, question, why do you condition on $V=v$? Isnt it a constant $a$?
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:59










  • $begingroup$
    Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
    $endgroup$
    – Song
    Dec 16 '18 at 19:09














6












6








6





$begingroup$

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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 4:02









SongSong

18.5k21651




18.5k21651












  • $begingroup$
    It's very nice!! Thank you!!
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:58










  • $begingroup$
    but, question, why do you condition on $V=v$? Isnt it a constant $a$?
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:59










  • $begingroup$
    Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
    $endgroup$
    – Song
    Dec 16 '18 at 19:09


















  • $begingroup$
    It's very nice!! Thank you!!
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:58










  • $begingroup$
    but, question, why do you condition on $V=v$? Isnt it a constant $a$?
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 18:59










  • $begingroup$
    Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
    $endgroup$
    – Song
    Dec 16 '18 at 19:09
















$begingroup$
It's very nice!! Thank you!!
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 18:58




$begingroup$
It's very nice!! Thank you!!
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 18:58












$begingroup$
but, question, why do you condition on $V=v$? Isnt it a constant $a$?
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 18:59




$begingroup$
but, question, why do you condition on $V=v$? Isnt it a constant $a$?
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 18:59












$begingroup$
Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
$endgroup$
– Song
Dec 16 '18 at 19:09




$begingroup$
Oh, it's my mistake. Maybe it was for notational consistency. But it may not cause problem because one can read $v$ appearing in all the expressions as $a$.
$endgroup$
– Song
Dec 16 '18 at 19:09


















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