Computing conditional expectation of independent uniform rv
$begingroup$
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
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
add a comment |
$begingroup$
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
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
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
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
edited Jan 14 at 19:21
Ethan Bolker
45.4k553120
45.4k553120
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
asked Dec 16 '18 at 2:19
Jimmy SabaterJimmy Sabater
2,560325
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
$endgroup$
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
$$
frac{1}{2-v}int_{{0<u<1,; v-1<u<v}} u^2 du = frac{1}{2-v}int_{v-1}^1 u^2 du = frac{1}{2-v}frac{u^3}{3}Big|^{u=1}_{u=v-1} = frac{v^2-v+1}{3}.
$$
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
answered Dec 16 '18 at 4:02
SongSong
18.5k21651
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
@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