Have I shown $operatorname{Var} X < infty iff mathbb E[X^2] < infty$?
Show that $operatorname{Var}{X} < infty iff mathbb E[X^2] < infty$
I am attempting to use the above without the fact that $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ if $mathbb E[X] < infty$, since we have not been given that $mathbb E[X] < infty$. (although do $operatorname{Var} X < infty$ as well as $mathbb E[X^2]< infty$ respectively imply that $mathbb E[X] < infty$ ?)
"$Rightarrow$"
$operatorname{Var} X= mathbb E[(X-mathbb E[X])^2]=int (X-mathbb E[X])^2dP=int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP$
Since $operatorname{Var} X < infty$, it is well-defined as an integral and therefore $int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP<infty$ implies $int X^2dP<infty$ so $mathbb E[X^2]<infty$
"$Leftarrow$" If $mathbb E[X^2] < infty$ implies that $mathbb E[X] < infty$ then it's easy game. Since $operatorname{Var} X = mathbb E[X^2]- mathbb E[X]^2$ which is $< infty$
Any pointers, improvements, pointing out of mistakes would be greatly appreciated.
probability-theory lebesgue-integral variance expected-value
edited Jan 4 at 12:05
Bernard
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
add a comment |
Show that $operatorname{Var}{X} < infty iff mathbb E[X^2] < infty$
I am attempting to use the above without the fact that $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ if $mathbb E[X] < infty$, since we have not been given that $mathbb E[X] < infty$. (although do $operatorname{Var} X < infty$ as well as $mathbb E[X^2]< infty$ respectively imply that $mathbb E[X] < infty$ ?)
"$Rightarrow$"
$operatorname{Var} X= mathbb E[(X-mathbb E[X])^2]=int (X-mathbb E[X])^2dP=int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP$
Since $operatorname{Var} X < infty$, it is well-defined as an integral and therefore $int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP<infty$ implies $int X^2dP<infty$ so $mathbb E[X^2]<infty$
"$Leftarrow$" If $mathbb E[X^2] < infty$ implies that $mathbb E[X] < infty$ then it's easy game. Since $operatorname{Var} X = mathbb E[X^2]- mathbb E[X]^2$ which is $< infty$
Any pointers, improvements, pointing out of mistakes would be greatly appreciated.
probability-theory lebesgue-integral variance expected-value
edited Jan 4 at 12:05
Bernard
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
add a comment |
Show that $operatorname{Var}{X} < infty iff mathbb E[X^2] < infty$
I am attempting to use the above without the fact that $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ if $mathbb E[X] < infty$, since we have not been given that $mathbb E[X] < infty$. (although do $operatorname{Var} X < infty$ as well as $mathbb E[X^2]< infty$ respectively imply that $mathbb E[X] < infty$ ?)
"$Rightarrow$"
$operatorname{Var} X= mathbb E[(X-mathbb E[X])^2]=int (X-mathbb E[X])^2dP=int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP$
Since $operatorname{Var} X < infty$, it is well-defined as an integral and therefore $int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP<infty$ implies $int X^2dP<infty$ so $mathbb E[X^2]<infty$
"$Leftarrow$" If $mathbb E[X^2] < infty$ implies that $mathbb E[X] < infty$ then it's easy game. Since $operatorname{Var} X = mathbb E[X^2]- mathbb E[X]^2$ which is $< infty$
Any pointers, improvements, pointing out of mistakes would be greatly appreciated.
probability-theory lebesgue-integral variance expected-value
edited Jan 4 at 12:05
Bernard
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
Show that $operatorname{Var}{X} < infty iff mathbb E[X^2] < infty$
I am attempting to use the above without the fact that $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ if $mathbb E[X] < infty$, since we have not been given that $mathbb E[X] < infty$. (although do $operatorname{Var} X < infty$ as well as $mathbb E[X^2]< infty$ respectively imply that $mathbb E[X] < infty$ ?)
"$Rightarrow$"
$operatorname{Var} X= mathbb E[(X-mathbb E[X])^2]=int (X-mathbb E[X])^2dP=int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP$
Since $operatorname{Var} X < infty$, it is well-defined as an integral and therefore $int(X^2-2Xmathbb E[X]+mathbb E[X]^2)dP<infty$ implies $int X^2dP<infty$ so $mathbb E[X^2]<infty$
"$Leftarrow$" If $mathbb E[X^2] < infty$ implies that $mathbb E[X] < infty$ then it's easy game. Since $operatorname{Var} X = mathbb E[X^2]- mathbb E[X]^2$ which is $< infty$
Any pointers, improvements, pointing out of mistakes would be greatly appreciated.
probability-theory lebesgue-integral variance expected-value
probability-theory lebesgue-integral variance expected-value
edited Jan 4 at 12:05
Bernard
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
edited Jan 4 at 12:05
Bernard
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
edited Jan 4 at 12:05
Bernard
118k639112
edited Jan 4 at 12:05
Bernard
118k639112
edited Jan 4 at 12:05
Bernard
118k639112
118k639112
asked Jan 4 at 9:57
SABOYSABOY
508311
asked Jan 4 at 9:57
SABOYSABOY
508311
asked Jan 4 at 9:57
SABOYSABOY
508311
508311
add a comment |
add a comment |
1 Answer
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To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $pm infty$. If $EX= pminfty$ then $(X-EX)^{2}=infty$ almost surely so variance is $infty$. Hence $var(X) <infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <infty$ if the variance is finite.
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
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To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $pm infty$. If $EX= pminfty$ then $(X-EX)^{2}=infty$ almost surely so variance is $infty$. Hence $var(X) <infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <infty$ if the variance is finite.
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $pm infty$. If $EX= pminfty$ then $(X-EX)^{2}=infty$ almost surely so variance is $infty$. Hence $var(X) <infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <infty$ if the variance is finite.
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $pm infty$. If $EX= pminfty$ then $(X-EX)^{2}=infty$ almost surely so variance is $infty$. Hence $var(X) <infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <infty$ if the variance is finite.
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $pm infty$. If $EX= pminfty$ then $(X-EX)^{2}=infty$ almost surely so variance is $infty$. Hence $var(X) <infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <infty$ if the variance is finite.
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
edited Jan 4 at 10:22
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Jan 4 at 10:13
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
add a comment |
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