Mixing
If $E[X]=int_{0}^{infty}(1-F_X(t)-F_X(-t))dt$ then why $E[X^2]=int_{0}^{infty}(1-F_{X^2}(t))dt$
$begingroup$
I know that from definition: $E[X]=int_{0}^{infty}(1-F_X(t)-F_X(-t))dt$
But I encountered the following claim:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t))dt$
Why isn't it the same as above? hence:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t)-F_{X^2}(-t))dt$
probability expected-value
asked Feb 1 at 8:37
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
I know that from definition: $E[X]=int_{0}^{infty}(1-F_X(t)-F_X(-t))dt$
But I encountered the following claim:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t))dt$
Why isn't it the same as above? hence:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t)-F_{X^2}(-t))dt$
probability expected-value
asked Feb 1 at 8:37
superuser123superuser123
48628
$endgroup$
$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43
add a comment |
$begingroup$
I know that from definition: $E[X]=int_{0}^{infty}(1-F_X(t)-F_X(-t))dt$
But I encountered the following claim:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t))dt$
Why isn't it the same as above? hence:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t)-F_{X^2}(-t))dt$
probability expected-value
asked Feb 1 at 8:37
superuser123superuser123
48628
$endgroup$
I know that from definition: $E[X]=int_{0}^{infty}(1-F_X(t)-F_X(-t))dt$
But I encountered the following claim:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t))dt$
Why isn't it the same as above? hence:
$E[X^2]=int_{0}^{infty}(1-F_{X^2}(t)-F_{X^2}(-t))dt$
probability expected-value
probability expected-value
asked Feb 1 at 8:37
superuser123superuser123
48628
asked Feb 1 at 8:37
superuser123superuser123
48628
edited Feb 1 at 9:27
superuser123
asked Feb 1 at 8:37
superuser123superuser123
48628
asked Feb 1 at 8:37
superuser123superuser123
48628
asked Feb 1 at 8:37
superuser123superuser123
48628
48628
$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43
add a comment |
$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43
$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43
$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43
add a comment |
1 Answer
1
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$begingroup$
They are actually the same, but because $X^2$ is always non negative $F_{X^2}(-t)=0$ and the expression simplifies to the one given.
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
They are actually the same, but because $X^2$ is always non negative $F_{X^2}(-t)=0$ and the expression simplifies to the one given.
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
They are actually the same, but because $X^2$ is always non negative $F_{X^2}(-t)=0$ and the expression simplifies to the one given.
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
They are actually the same, but because $X^2$ is always non negative $F_{X^2}(-t)=0$ and the expression simplifies to the one given.
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
$endgroup$
They are actually the same, but because $X^2$ is always non negative $F_{X^2}(-t)=0$ and the expression simplifies to the one given.
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
answered Feb 1 at 8:40
b00n heTb00n heT
10.5k12335
10.5k12335
add a comment |
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$begingroup$
The last expression should read $E[X^color{red}{2}]=...$
$endgroup$
– Math-fun
Feb 1 at 8:43