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does $E(XY) = E(XX)$ if $X$ and $Y$ are i.i.d standard normal?
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If I have two random variables $X$ and $Y$ which are independent and identically distributed (i.i.d) as Standard normal $sim N(0,1)$
does $E(X^2) = E(XY)$ since $X$ and $Y$ are i.i.d?
please explain it step by step, I'm a beginner in probability. and if it is true, is it true for any pair of i.i.d random variable no matter what distribution it is (provided that it has a finite mean)?
probability probability-distributions expectation
edited Jul 24 at 0:41
BDN
573417
asked Jul 24 at 0:13
user3270418
374212
add a comment |Â
up vote
2
down vote
favorite
If I have two random variables $X$ and $Y$ which are independent and identically distributed (i.i.d) as Standard normal $sim N(0,1)$
does $E(X^2) = E(XY)$ since $X$ and $Y$ are i.i.d?
please explain it step by step, I'm a beginner in probability. and if it is true, is it true for any pair of i.i.d random variable no matter what distribution it is (provided that it has a finite mean)?
probability probability-distributions expectation
edited Jul 24 at 0:41
BDN
573417
asked Jul 24 at 0:13
user3270418
374212
2
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If I have two random variables $X$ and $Y$ which are independent and identically distributed (i.i.d) as Standard normal $sim N(0,1)$
does $E(X^2) = E(XY)$ since $X$ and $Y$ are i.i.d?
please explain it step by step, I'm a beginner in probability. and if it is true, is it true for any pair of i.i.d random variable no matter what distribution it is (provided that it has a finite mean)?
probability probability-distributions expectation
edited Jul 24 at 0:41
BDN
573417
asked Jul 24 at 0:13
user3270418
374212
If I have two random variables $X$ and $Y$ which are independent and identically distributed (i.i.d) as Standard normal $sim N(0,1)$
does $E(X^2) = E(XY)$ since $X$ and $Y$ are i.i.d?
please explain it step by step, I'm a beginner in probability. and if it is true, is it true for any pair of i.i.d random variable no matter what distribution it is (provided that it has a finite mean)?
probability probability-distributions expectation
edited Jul 24 at 0:41
BDN
573417
asked Jul 24 at 0:13
user3270418
374212
edited Jul 24 at 0:41
BDN
573417
edited Jul 24 at 0:41
BDN
573417
edited Jul 24 at 0:41
BDN
573417
573417
asked Jul 24 at 0:13
user3270418
374212
asked Jul 24 at 0:13
user3270418
374212
asked Jul 24 at 0:13
user3270418
374212
374212
2
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30
add a comment |Â
2
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30
2
2
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
No.
Since $X$ has variance $1$ and expectation $0$
$$mathbbE[X^2] =mathbbE[X^2] - mathbbE[X]^2 = operatornameVar[X] = 1$$ by definition, while, since $X,Y$ are independent,
$$
mathbbE[XY] = mathbbE[X]mathbbE[Y]=0cdot 0 = 0,.
$$
answered Jul 24 at 0:18
Clement C.
47.1k33682
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
No.
Since $X$ has variance $1$ and expectation $0$
$$mathbbE[X^2] =mathbbE[X^2] - mathbbE[X]^2 = operatornameVar[X] = 1$$ by definition, while, since $X,Y$ are independent,
$$
mathbbE[XY] = mathbbE[X]mathbbE[Y]=0cdot 0 = 0,.
$$
answered Jul 24 at 0:18
Clement C.
47.1k33682
add a comment |Â
up vote
6
down vote
accepted
No.
Since $X$ has variance $1$ and expectation $0$
$$mathbbE[X^2] =mathbbE[X^2] - mathbbE[X]^2 = operatornameVar[X] = 1$$ by definition, while, since $X,Y$ are independent,
$$
mathbbE[XY] = mathbbE[X]mathbbE[Y]=0cdot 0 = 0,.
$$
answered Jul 24 at 0:18
Clement C.
47.1k33682
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
No.
Since $X$ has variance $1$ and expectation $0$
$$mathbbE[X^2] =mathbbE[X^2] - mathbbE[X]^2 = operatornameVar[X] = 1$$ by definition, while, since $X,Y$ are independent,
$$
mathbbE[XY] = mathbbE[X]mathbbE[Y]=0cdot 0 = 0,.
$$
answered Jul 24 at 0:18
Clement C.
47.1k33682
No.
Since $X$ has variance $1$ and expectation $0$
$$mathbbE[X^2] =mathbbE[X^2] - mathbbE[X]^2 = operatornameVar[X] = 1$$ by definition, while, since $X,Y$ are independent,
$$
mathbbE[XY] = mathbbE[X]mathbbE[Y]=0cdot 0 = 0,.
$$
answered Jul 24 at 0:18
Clement C.
47.1k33682
answered Jul 24 at 0:18
Clement C.
47.1k33682
answered Jul 24 at 0:18
Clement C.
47.1k33682
answered Jul 24 at 0:18
Clement C.
47.1k33682
47.1k33682
add a comment |Â
add a comment |Â
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2
By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$.
– GEdgar
Jul 24 at 0:24
@GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $operatornameVar X =0$, i.e., $X$ is almost surely (a.s.) constant
– Clement C.
Jul 24 at 0:30
I think the simplest proof would be that x^2 can never be negative where as xy can be.
– barrycarter
Jul 24 at 1:30