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
asked 4 hours ago
Jimmy Sabater
1,846218
add a comment |
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
asked 4 hours ago
Jimmy Sabater
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
add a comment |
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
asked 4 hours ago
Jimmy Sabater
1,846218
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
asked 4 hours ago
Jimmy Sabater
1,846218
asked 4 hours ago
Jimmy Sabater
1,846218
asked 4 hours ago
Jimmy Sabater
1,846218
asked 4 hours ago
Jimmy Sabater
1,846218
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
add a comment |
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
add a comment |
1 Answer
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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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered 2 hours ago
Song
2,965214
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered 2 hours ago
Song
2,965214
add a comment |
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered 2 hours ago
Song
2,965214
add a comment |
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered 2 hours ago
Song
2,965214
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered 2 hours ago
Song
2,965214
answered 2 hours ago
Song
2,965214
answered 2 hours ago
Song
2,965214
answered 2 hours ago
Song
2,965214
2,965214
add a comment |
add a comment |
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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