How to use $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ on $operatorname{Var}(X)=0$
$begingroup$
In our lecture, we have proven that
$mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ (*)
then we go on say if $operatorname{Var}(X)=0$ it follows using (*) that:
$mathbb P((X-mathbb E[X])^2=0)=1$ and therefore $mathbb P(X=mathbb E[X])=1$
I have seen other proofs via contradiction to prove $mathbb P(X=mathbb E[X])=1$ but all have been without the use of (*). As my professor makes direct reference to $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$, I assume it should be somewhat easier. Nonetheless, I do not understand how it becomes simpler. Any help is greatly appreciated
probability-theory stochastic-processes variance expected-value
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
In our lecture, we have proven that
$mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ (*)
then we go on say if $operatorname{Var}(X)=0$ it follows using (*) that:
$mathbb P((X-mathbb E[X])^2=0)=1$ and therefore $mathbb P(X=mathbb E[X])=1$
I have seen other proofs via contradiction to prove $mathbb P(X=mathbb E[X])=1$ but all have been without the use of (*). As my professor makes direct reference to $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$, I assume it should be somewhat easier. Nonetheless, I do not understand how it becomes simpler. Any help is greatly appreciated
probability-theory stochastic-processes variance expected-value
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
$endgroup$
2
$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43
add a comment |
$begingroup$
In our lecture, we have proven that
$mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ (*)
then we go on say if $operatorname{Var}(X)=0$ it follows using (*) that:
$mathbb P((X-mathbb E[X])^2=0)=1$ and therefore $mathbb P(X=mathbb E[X])=1$
I have seen other proofs via contradiction to prove $mathbb P(X=mathbb E[X])=1$ but all have been without the use of (*). As my professor makes direct reference to $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$, I assume it should be somewhat easier. Nonetheless, I do not understand how it becomes simpler. Any help is greatly appreciated
probability-theory stochastic-processes variance expected-value
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
$endgroup$
In our lecture, we have proven that
$mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$ (*)
then we go on say if $operatorname{Var}(X)=0$ it follows using (*) that:
$mathbb P((X-mathbb E[X])^2=0)=1$ and therefore $mathbb P(X=mathbb E[X])=1$
I have seen other proofs via contradiction to prove $mathbb P(X=mathbb E[X])=1$ but all have been without the use of (*). As my professor makes direct reference to $mathbb E[(X-mathbb E[X])^2]=mathbb E[X^2]-mathbb E[X]^2$, I assume it should be somewhat easier. Nonetheless, I do not understand how it becomes simpler. Any help is greatly appreciated
probability-theory stochastic-processes variance expected-value
probability-theory stochastic-processes variance expected-value
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
asked Dec 28 '18 at 23:35
SABOYSABOY
656311
656311
2
$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43
add a comment |
2
$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43
2
2
$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43
$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43
add a comment |
1 Answer
1
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oldest
votes
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The proof uses the following fact. Namely, if $EY=0$ where $Ygeq 0$, then $Y=0$ with probability one. Indeed by markov's inequality $P(Y> 1/n)leq nEY=0$ so whence by measure continuity $P(Y>0)=lim_{nto infty}P(Y>1/n)=0$. Thus $P(Y=0)=1$.
In your proof use the above fact with $Y=(X-EX)^2$. Then we have that $(X-EX)^2=0$ with probability one i.e. $X=EX$ with probability one.
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
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add a comment |
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$begingroup$
The proof uses the following fact. Namely, if $EY=0$ where $Ygeq 0$, then $Y=0$ with probability one. Indeed by markov's inequality $P(Y> 1/n)leq nEY=0$ so whence by measure continuity $P(Y>0)=lim_{nto infty}P(Y>1/n)=0$. Thus $P(Y=0)=1$.
In your proof use the above fact with $Y=(X-EX)^2$. Then we have that $(X-EX)^2=0$ with probability one i.e. $X=EX$ with probability one.
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
The proof uses the following fact. Namely, if $EY=0$ where $Ygeq 0$, then $Y=0$ with probability one. Indeed by markov's inequality $P(Y> 1/n)leq nEY=0$ so whence by measure continuity $P(Y>0)=lim_{nto infty}P(Y>1/n)=0$. Thus $P(Y=0)=1$.
In your proof use the above fact with $Y=(X-EX)^2$. Then we have that $(X-EX)^2=0$ with probability one i.e. $X=EX$ with probability one.
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
The proof uses the following fact. Namely, if $EY=0$ where $Ygeq 0$, then $Y=0$ with probability one. Indeed by markov's inequality $P(Y> 1/n)leq nEY=0$ so whence by measure continuity $P(Y>0)=lim_{nto infty}P(Y>1/n)=0$. Thus $P(Y=0)=1$.
In your proof use the above fact with $Y=(X-EX)^2$. Then we have that $(X-EX)^2=0$ with probability one i.e. $X=EX$ with probability one.
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
The proof uses the following fact. Namely, if $EY=0$ where $Ygeq 0$, then $Y=0$ with probability one. Indeed by markov's inequality $P(Y> 1/n)leq nEY=0$ so whence by measure continuity $P(Y>0)=lim_{nto infty}P(Y>1/n)=0$. Thus $P(Y=0)=1$.
In your proof use the above fact with $Y=(X-EX)^2$. Then we have that $(X-EX)^2=0$ with probability one i.e. $X=EX$ with probability one.
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
answered Dec 28 '18 at 23:57
Foobaz JohnFoobaz John
22.2k41452
22.2k41452
add a comment |
add a comment |
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$begingroup$
What is your definition of variance? I have trouble seeing how the right-hand side of your (*) is relevant for your task.
$endgroup$
– Henning Makholm
Dec 28 '18 at 23:43