$Xin L^2$ implies $Y:=(X-E[X])^2 in L^1$
0
$begingroup$
Given the random variable $X in L^2$, we define a new variable $Y:=(X-E[X])^2$, where $E$ is the expectation. Why can we conclude that $Yin L^1$?
Is it because for any function $f in L^2$ the norm is $(int_X|f|^2)^{frac{1}{2}}$, so the norm squared would be a new function $g in L^2$ with $(int_X|g|)$?
probability-theory measure-theory lebesgue-integral
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
$endgroup$
add a comment |
0
$begingroup$
Given the random variable $X in L^2$, we define a new variable $Y:=(X-E[X])^2$, where $E$ is the expectation. Why can we conclude that $Yin L^1$?
Is it because for any function $f in L^2$ the norm is $(int_X|f|^2)^{frac{1}{2}}$, so the norm squared would be a new function $g in L^2$ with $(int_X|g|)$?
probability-theory measure-theory lebesgue-integral
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
$endgroup$
$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12
add a comment |
0
0
0
$begingroup$
Given the random variable $X in L^2$, we define a new variable $Y:=(X-E[X])^2$, where $E$ is the expectation. Why can we conclude that $Yin L^1$?
Is it because for any function $f in L^2$ the norm is $(int_X|f|^2)^{frac{1}{2}}$, so the norm squared would be a new function $g in L^2$ with $(int_X|g|)$?
probability-theory measure-theory lebesgue-integral
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
$endgroup$
Given the random variable $X in L^2$, we define a new variable $Y:=(X-E[X])^2$, where $E$ is the expectation. Why can we conclude that $Yin L^1$?
Is it because for any function $f in L^2$ the norm is $(int_X|f|^2)^{frac{1}{2}}$, so the norm squared would be a new function $g in L^2$ with $(int_X|g|)$?
probability-theory measure-theory lebesgue-integral
probability-theory measure-theory lebesgue-integral
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
asked Dec 19 '18 at 19:16
ThesinusThesinus
254210
254210
$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12
add a comment |
$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12
$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12
add a comment |
1 Answer
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$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 le E[X^2] < infty.$$
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 le E[X^2] < infty.$$
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
add a comment |
1
$begingroup$
$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 le E[X^2] < infty.$$
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
add a comment |
1
1
1
$begingroup$
$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 le E[X^2] < infty.$$
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
$endgroup$
$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 le E[X^2] < infty.$$
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
answered Dec 19 '18 at 19:24
angryavianangryavian
41.8k23381
41.8k23381
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
add a comment |
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
How do you actually get$E[(X - E[X])^2] = E[X^2] - E[X]^2$?
$endgroup$
– Thesinus
Jan 17 at 16:28
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
$begingroup$
@Thesinus $E[(X-E[X])^2] = E[X^2 - 2 X E[X] + (E[X])^2] = E[X^2] - 2 (E[X])^2 + (E[X])^2$.
$endgroup$
– angryavian
Jan 17 at 16:59
add a comment |
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$begingroup$
Just expand that square and use that if $Y in L^2$ then $Y in L^1$ (this is true on probability spaces, using Holder's inequality). (You'll also use that the sum of $L^1$ functions is$L^1$)
$endgroup$
– Lorenzo
Dec 19 '18 at 19:19
$begingroup$
By definition, $X$ is square integrable if its square is integrable, hence so is $Y.$ (I am not sure if you are asking for this tautology or something else...)
$endgroup$
– Will M.
Dec 19 '18 at 20:12