Mixing
Proof of $ E(XY) = E(X) E(Y) $
$begingroup$
When two random variables are statistically independent, the
expectation of their product is the product of their expectations.
I found this on wikipedia :
https://en.wikipedia.org/wiki/Product_distribution
$$ E(XY) = E(X) E(Y) $$
Nevertheless, I can't find a simple proof. I think that in my probability books, they are skipping a few steps writting :
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} xy P(X = x) P(Y=y) $$
I don't understand this equality because I was thinking that
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} z P(XY = z) $$
is true (i guess), and I don't know how to go from the second equality to the first one.
The link to the wikipedia page gives another proof using conditional expectency, which is something I'm rather unfamiliar with. Could you please give me a simple explanation/proof of the link between the two Latex lines ?
Thank you !
probability probability-theory expected-value
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
$endgroup$
add a comment |
$begingroup$
When two random variables are statistically independent, the
expectation of their product is the product of their expectations.
I found this on wikipedia :
https://en.wikipedia.org/wiki/Product_distribution
$$ E(XY) = E(X) E(Y) $$
Nevertheless, I can't find a simple proof. I think that in my probability books, they are skipping a few steps writting :
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} xy P(X = x) P(Y=y) $$
I don't understand this equality because I was thinking that
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} z P(XY = z) $$
is true (i guess), and I don't know how to go from the second equality to the first one.
The link to the wikipedia page gives another proof using conditional expectency, which is something I'm rather unfamiliar with. Could you please give me a simple explanation/proof of the link between the two Latex lines ?
Thank you !
probability probability-theory expected-value
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
$endgroup$
$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19
add a comment |
$begingroup$
When two random variables are statistically independent, the
expectation of their product is the product of their expectations.
I found this on wikipedia :
https://en.wikipedia.org/wiki/Product_distribution
$$ E(XY) = E(X) E(Y) $$
Nevertheless, I can't find a simple proof. I think that in my probability books, they are skipping a few steps writting :
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} xy P(X = x) P(Y=y) $$
I don't understand this equality because I was thinking that
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} z P(XY = z) $$
is true (i guess), and I don't know how to go from the second equality to the first one.
The link to the wikipedia page gives another proof using conditional expectency, which is something I'm rather unfamiliar with. Could you please give me a simple explanation/proof of the link between the two Latex lines ?
Thank you !
probability probability-theory expected-value
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
$endgroup$
When two random variables are statistically independent, the
expectation of their product is the product of their expectations.
I found this on wikipedia :
https://en.wikipedia.org/wiki/Product_distribution
$$ E(XY) = E(X) E(Y) $$
Nevertheless, I can't find a simple proof. I think that in my probability books, they are skipping a few steps writting :
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} xy P(X = x) P(Y=y) $$
I don't understand this equality because I was thinking that
$$ E(XY) = sum_{x in D_1 } sum_{y in D_2} z P(XY = z) $$
is true (i guess), and I don't know how to go from the second equality to the first one.
The link to the wikipedia page gives another proof using conditional expectency, which is something I'm rather unfamiliar with. Could you please give me a simple explanation/proof of the link between the two Latex lines ?
Thank you !
probability probability-theory expected-value
probability probability-theory expected-value
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
asked Jan 29 at 7:43
Marine GalantinMarine Galantin
945419
945419
$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19
add a comment |
$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19
$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19
$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you do in your problem, we shall restrict ourselves to the discrete case, where $X$ and $Y$ are random variables taking at most countably infinitely many possible values. Let us call this set of values $S$. Let $T$ denote the set of products of two elements in $S$. Then $XY$ is a discrete random variable taking values in the set $T$.
Now suppose I ask you what the probability of $XY$ taking the value $z$ is. Well clearly if $X = x$, then $Y$ must equal $z/x$. And since $X$ must take some value, we must have that
$$
P(XY=z) = sum_{x in S} P(X = x text{ and } Y = z/x).
$$
Now, since $X$ and $Y$ are independent, we have that $P(X = x text{ and } Y = z/x) = P(X=x)P(y=z/x)$. Thus
$$
P(XY=z) = sum_{x in S} P(X = x) P(Y = z/x).
$$
By definition of expectation,
$$
E(XY) = sum_{z in T} zP(XY=z).
$$
Using our expression we just computed for $P(XY=z)$, we substitute
$$
E(XY) = sum_{z in T} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Note that if $z/x$ is not in our set $S$, $P(Y = z/x) = 0$, so we may simplify the above summation to
$$
E(XY) = sum_{z/x in S} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Now make the substitution $y = z/x$. Noting that $z = xcdot z/x = xy$, we conclude
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y).
$$
From here, we factor
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y) = left(sum_{x in S}xP(X=x) right)left(sum_{y in S}yP(Y=y) right) = E(X)E(Y),
$$
and the desired result has been obtained.
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
$endgroup$
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
add a comment |
$begingroup$
Here is a rough outline (not very rigorous, perhaps) -
$$mathbb{E}[XY] = sum_z zP(XY=z)$$
$$ = sum_{y,z} zP(X=z/y|Y=y)P(Y=y) $$
$$= sum_{y,z} zP(X = z/y)P(Y=y)$$
$$= sum_{x,y:xy=z} xyP(X=x)P(Y=y)$$
From the 2nd to the 3rd line, I used the independence of $X$ and $Y$ to claim that $P(X=a|Y=b) = P(X=a)$.
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you do in your problem, we shall restrict ourselves to the discrete case, where $X$ and $Y$ are random variables taking at most countably infinitely many possible values. Let us call this set of values $S$. Let $T$ denote the set of products of two elements in $S$. Then $XY$ is a discrete random variable taking values in the set $T$.
Now suppose I ask you what the probability of $XY$ taking the value $z$ is. Well clearly if $X = x$, then $Y$ must equal $z/x$. And since $X$ must take some value, we must have that
$$
P(XY=z) = sum_{x in S} P(X = x text{ and } Y = z/x).
$$
Now, since $X$ and $Y$ are independent, we have that $P(X = x text{ and } Y = z/x) = P(X=x)P(y=z/x)$. Thus
$$
P(XY=z) = sum_{x in S} P(X = x) P(Y = z/x).
$$
By definition of expectation,
$$
E(XY) = sum_{z in T} zP(XY=z).
$$
Using our expression we just computed for $P(XY=z)$, we substitute
$$
E(XY) = sum_{z in T} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Note that if $z/x$ is not in our set $S$, $P(Y = z/x) = 0$, so we may simplify the above summation to
$$
E(XY) = sum_{z/x in S} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Now make the substitution $y = z/x$. Noting that $z = xcdot z/x = xy$, we conclude
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y).
$$
From here, we factor
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y) = left(sum_{x in S}xP(X=x) right)left(sum_{y in S}yP(Y=y) right) = E(X)E(Y),
$$
and the desired result has been obtained.
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
$endgroup$
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
add a comment |
$begingroup$
As you do in your problem, we shall restrict ourselves to the discrete case, where $X$ and $Y$ are random variables taking at most countably infinitely many possible values. Let us call this set of values $S$. Let $T$ denote the set of products of two elements in $S$. Then $XY$ is a discrete random variable taking values in the set $T$.
Now suppose I ask you what the probability of $XY$ taking the value $z$ is. Well clearly if $X = x$, then $Y$ must equal $z/x$. And since $X$ must take some value, we must have that
$$
P(XY=z) = sum_{x in S} P(X = x text{ and } Y = z/x).
$$
Now, since $X$ and $Y$ are independent, we have that $P(X = x text{ and } Y = z/x) = P(X=x)P(y=z/x)$. Thus
$$
P(XY=z) = sum_{x in S} P(X = x) P(Y = z/x).
$$
By definition of expectation,
$$
E(XY) = sum_{z in T} zP(XY=z).
$$
Using our expression we just computed for $P(XY=z)$, we substitute
$$
E(XY) = sum_{z in T} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Note that if $z/x$ is not in our set $S$, $P(Y = z/x) = 0$, so we may simplify the above summation to
$$
E(XY) = sum_{z/x in S} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Now make the substitution $y = z/x$. Noting that $z = xcdot z/x = xy$, we conclude
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y).
$$
From here, we factor
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y) = left(sum_{x in S}xP(X=x) right)left(sum_{y in S}yP(Y=y) right) = E(X)E(Y),
$$
and the desired result has been obtained.
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
$endgroup$
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
add a comment |
$begingroup$
As you do in your problem, we shall restrict ourselves to the discrete case, where $X$ and $Y$ are random variables taking at most countably infinitely many possible values. Let us call this set of values $S$. Let $T$ denote the set of products of two elements in $S$. Then $XY$ is a discrete random variable taking values in the set $T$.
Now suppose I ask you what the probability of $XY$ taking the value $z$ is. Well clearly if $X = x$, then $Y$ must equal $z/x$. And since $X$ must take some value, we must have that
$$
P(XY=z) = sum_{x in S} P(X = x text{ and } Y = z/x).
$$
Now, since $X$ and $Y$ are independent, we have that $P(X = x text{ and } Y = z/x) = P(X=x)P(y=z/x)$. Thus
$$
P(XY=z) = sum_{x in S} P(X = x) P(Y = z/x).
$$
By definition of expectation,
$$
E(XY) = sum_{z in T} zP(XY=z).
$$
Using our expression we just computed for $P(XY=z)$, we substitute
$$
E(XY) = sum_{z in T} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Note that if $z/x$ is not in our set $S$, $P(Y = z/x) = 0$, so we may simplify the above summation to
$$
E(XY) = sum_{z/x in S} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Now make the substitution $y = z/x$. Noting that $z = xcdot z/x = xy$, we conclude
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y).
$$
From here, we factor
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y) = left(sum_{x in S}xP(X=x) right)left(sum_{y in S}yP(Y=y) right) = E(X)E(Y),
$$
and the desired result has been obtained.
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
$endgroup$
As you do in your problem, we shall restrict ourselves to the discrete case, where $X$ and $Y$ are random variables taking at most countably infinitely many possible values. Let us call this set of values $S$. Let $T$ denote the set of products of two elements in $S$. Then $XY$ is a discrete random variable taking values in the set $T$.
Now suppose I ask you what the probability of $XY$ taking the value $z$ is. Well clearly if $X = x$, then $Y$ must equal $z/x$. And since $X$ must take some value, we must have that
$$
P(XY=z) = sum_{x in S} P(X = x text{ and } Y = z/x).
$$
Now, since $X$ and $Y$ are independent, we have that $P(X = x text{ and } Y = z/x) = P(X=x)P(y=z/x)$. Thus
$$
P(XY=z) = sum_{x in S} P(X = x) P(Y = z/x).
$$
By definition of expectation,
$$
E(XY) = sum_{z in T} zP(XY=z).
$$
Using our expression we just computed for $P(XY=z)$, we substitute
$$
E(XY) = sum_{z in T} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Note that if $z/x$ is not in our set $S$, $P(Y = z/x) = 0$, so we may simplify the above summation to
$$
E(XY) = sum_{z/x in S} sum_{x in S} zP(X=x)P(Y=z/x).
$$
Now make the substitution $y = z/x$. Noting that $z = xcdot z/x = xy$, we conclude
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y).
$$
From here, we factor
$$
E(XY) = sum_{y in S} sum_{x in S} xyP(X=x)P(Y=y) = left(sum_{x in S}xP(X=x) right)left(sum_{y in S}yP(Y=y) right) = E(X)E(Y),
$$
and the desired result has been obtained.
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
answered Jan 29 at 7:57
eepperly16eepperly16
3,06111125
3,06111125
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
add a comment |
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
that's great, thank you. So i was right thinking it was not that obvious to have the mentionned equality? Or am I just over thinking things?
$endgroup$
– Marine Galantin
Jan 29 at 8:01
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
@MarineGalantin Not obvious to have which equality? The first or second in your original post, or some equality from my answer?
$endgroup$
– eepperly16
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
$begingroup$
the first one I mentionned, the one you proved :)
$endgroup$
– Marine Galantin
Jan 29 at 8:03
1
1
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
$begingroup$
@MarineGalantin I don't think it's obvious at all. In fact, for random variables which are not discrete, a formal, rigorous proof requires a bit of trickiness
$endgroup$
– eepperly16
Jan 29 at 8:07
add a comment |
$begingroup$
Here is a rough outline (not very rigorous, perhaps) -
$$mathbb{E}[XY] = sum_z zP(XY=z)$$
$$ = sum_{y,z} zP(X=z/y|Y=y)P(Y=y) $$
$$= sum_{y,z} zP(X = z/y)P(Y=y)$$
$$= sum_{x,y:xy=z} xyP(X=x)P(Y=y)$$
From the 2nd to the 3rd line, I used the independence of $X$ and $Y$ to claim that $P(X=a|Y=b) = P(X=a)$.
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
Here is a rough outline (not very rigorous, perhaps) -
$$mathbb{E}[XY] = sum_z zP(XY=z)$$
$$ = sum_{y,z} zP(X=z/y|Y=y)P(Y=y) $$
$$= sum_{y,z} zP(X = z/y)P(Y=y)$$
$$= sum_{x,y:xy=z} xyP(X=x)P(Y=y)$$
From the 2nd to the 3rd line, I used the independence of $X$ and $Y$ to claim that $P(X=a|Y=b) = P(X=a)$.
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
Here is a rough outline (not very rigorous, perhaps) -
$$mathbb{E}[XY] = sum_z zP(XY=z)$$
$$ = sum_{y,z} zP(X=z/y|Y=y)P(Y=y) $$
$$= sum_{y,z} zP(X = z/y)P(Y=y)$$
$$= sum_{x,y:xy=z} xyP(X=x)P(Y=y)$$
From the 2nd to the 3rd line, I used the independence of $X$ and $Y$ to claim that $P(X=a|Y=b) = P(X=a)$.
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
$endgroup$
Here is a rough outline (not very rigorous, perhaps) -
$$mathbb{E}[XY] = sum_z zP(XY=z)$$
$$ = sum_{y,z} zP(X=z/y|Y=y)P(Y=y) $$
$$= sum_{y,z} zP(X = z/y)P(Y=y)$$
$$= sum_{x,y:xy=z} xyP(X=x)P(Y=y)$$
From the 2nd to the 3rd line, I used the independence of $X$ and $Y$ to claim that $P(X=a|Y=b) = P(X=a)$.
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 8:06
Aditya DuaAditya Dua
1,15418
1,15418
add a comment |
add a comment |
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$begingroup$
If the two variables are independent, then $E[Y|X]=E[Y]$; that is, $E[Y]$ is independent of $X$. Thus, $$ E[XY]stackrel{substack{text{conditional}\text{probability}}}= E[X,E[Y|X]]stackrel{text{independence}}= E[X,E[Y]]stackrel{substack{text{linearity of}\text{expectation}}}= E[X],E[Y] $$
$endgroup$
– robjohn♦
Jan 29 at 10:19