Prove / Disprove that $Ebig((X−a)^4big)$ is minimized when $a=E(X)$.
$begingroup$
Let $X$ be a random variable with moment up to order $4$. Prove / disprove that $Ebig((X−a)^4big)$ is minimized when $a=E(X)$.
I've already proved that $E((X−a)^2)$ is minimized as such, but I'm lost with the 4 power. I've tried things from differentiating to opening up the parenthesis and trying to work from there but nothing worked.
plus I couldn't come up with any counter-example.
any help is appreciated!
thanks
probability optimization random means expected-value
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
$endgroup$
add a comment |
$begingroup$
Let $X$ be a random variable with moment up to order $4$. Prove / disprove that $Ebig((X−a)^4big)$ is minimized when $a=E(X)$.
I've already proved that $E((X−a)^2)$ is minimized as such, but I'm lost with the 4 power. I've tried things from differentiating to opening up the parenthesis and trying to work from there but nothing worked.
plus I couldn't come up with any counter-example.
any help is appreciated!
thanks
probability optimization random means expected-value
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
$endgroup$
1
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08
add a comment |
$begingroup$
Let $X$ be a random variable with moment up to order $4$. Prove / disprove that $Ebig((X−a)^4big)$ is minimized when $a=E(X)$.
I've already proved that $E((X−a)^2)$ is minimized as such, but I'm lost with the 4 power. I've tried things from differentiating to opening up the parenthesis and trying to work from there but nothing worked.
plus I couldn't come up with any counter-example.
any help is appreciated!
thanks
probability optimization random means expected-value
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
$endgroup$
Let $X$ be a random variable with moment up to order $4$. Prove / disprove that $Ebig((X−a)^4big)$ is minimized when $a=E(X)$.
I've already proved that $E((X−a)^2)$ is minimized as such, but I'm lost with the 4 power. I've tried things from differentiating to opening up the parenthesis and trying to work from there but nothing worked.
plus I couldn't come up with any counter-example.
any help is appreciated!
thanks
probability optimization random means expected-value
probability optimization random means expected-value
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
edited Dec 14 '18 at 21:08
Batominovski
33.1k33293
33.1k33293
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
asked Dec 14 '18 at 21:04
Itay DayItay Day
111
111
1
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08
add a comment |
1
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08
1
1
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Expanding $mathbb{E}left[(X-a)^4right]$ yields
$$f(a):=mathbb{E}left[(X-a)^4right]=mathbb{E}[X^4]-4,mathbb{E}[X^3],a+6,mathbb{E}[X^2],a^2-4,mathbb{E}[X],a^3+a^4,.$$
Thus,
$$f'(a)=-4,mathbb{E}[X^3]+12,mathbb{E}[X^2],a-12,mathbb{E}[X],a^2+4,a^3,.$$
Therefore, if $a=mathbb{E}[X]$ minimizes $f$, then $f'big(mathbb{E}[X]big)=0$, and so
$$-mathbb{E}[X^3]+3,mathbb{E}[X^2],mathbb{E}[X]-2,big(mathbb{E}[X]big)^3=0,.tag{*}$$
Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $mathbb{E}[X]=0$ but $mathbb{E}[X^3]neq 0$.
Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$mathbb{P}[X=-1]=frac{2}{3}text{ and }mathbb{P}[X=2]=frac13,.$$ Then, $$mathbb{E}[X]=0text{ whereas }mathbb{E}[X^3]=2neq 0,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18left(sqrt[3]{2}-1right)neq 0=mathbb{E}[X],.$$
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
$endgroup$
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
add a comment |
$begingroup$
For a counter example take a look at a random variable with $Xin{1,-2}$ with probabilities $$P(X=+1) =frac{2}{3}, qquad P(X=-2) = frac{1}{3},.$$
Now, we can see that
$$E(X) =frac{2}{3} cdot 1 + frac{1}{3}cdot(-2) = 0,.$$
However,
$$E((X-a)^4) = frac{2}{3} cdot (1-a)^4 + frac{1}{3}cdot(-2-a)^4,.$$
We have
$$ frac{81}{16}= E((X+tfrac12)^4) < E(X^4) = 6$$
providing a counterexample.
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
$endgroup$
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039880%2fprove-disprove-that-e-bigx%25e2%2588%2592a4-big-is-minimized-when-a-ex%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Expanding $mathbb{E}left[(X-a)^4right]$ yields
$$f(a):=mathbb{E}left[(X-a)^4right]=mathbb{E}[X^4]-4,mathbb{E}[X^3],a+6,mathbb{E}[X^2],a^2-4,mathbb{E}[X],a^3+a^4,.$$
Thus,
$$f'(a)=-4,mathbb{E}[X^3]+12,mathbb{E}[X^2],a-12,mathbb{E}[X],a^2+4,a^3,.$$
Therefore, if $a=mathbb{E}[X]$ minimizes $f$, then $f'big(mathbb{E}[X]big)=0$, and so
$$-mathbb{E}[X^3]+3,mathbb{E}[X^2],mathbb{E}[X]-2,big(mathbb{E}[X]big)^3=0,.tag{*}$$
Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $mathbb{E}[X]=0$ but $mathbb{E}[X^3]neq 0$.
Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$mathbb{P}[X=-1]=frac{2}{3}text{ and }mathbb{P}[X=2]=frac13,.$$ Then, $$mathbb{E}[X]=0text{ whereas }mathbb{E}[X^3]=2neq 0,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18left(sqrt[3]{2}-1right)neq 0=mathbb{E}[X],.$$
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
$endgroup$
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
add a comment |
$begingroup$
Expanding $mathbb{E}left[(X-a)^4right]$ yields
$$f(a):=mathbb{E}left[(X-a)^4right]=mathbb{E}[X^4]-4,mathbb{E}[X^3],a+6,mathbb{E}[X^2],a^2-4,mathbb{E}[X],a^3+a^4,.$$
Thus,
$$f'(a)=-4,mathbb{E}[X^3]+12,mathbb{E}[X^2],a-12,mathbb{E}[X],a^2+4,a^3,.$$
Therefore, if $a=mathbb{E}[X]$ minimizes $f$, then $f'big(mathbb{E}[X]big)=0$, and so
$$-mathbb{E}[X^3]+3,mathbb{E}[X^2],mathbb{E}[X]-2,big(mathbb{E}[X]big)^3=0,.tag{*}$$
Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $mathbb{E}[X]=0$ but $mathbb{E}[X^3]neq 0$.
Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$mathbb{P}[X=-1]=frac{2}{3}text{ and }mathbb{P}[X=2]=frac13,.$$ Then, $$mathbb{E}[X]=0text{ whereas }mathbb{E}[X^3]=2neq 0,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18left(sqrt[3]{2}-1right)neq 0=mathbb{E}[X],.$$
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
$endgroup$
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
add a comment |
$begingroup$
Expanding $mathbb{E}left[(X-a)^4right]$ yields
$$f(a):=mathbb{E}left[(X-a)^4right]=mathbb{E}[X^4]-4,mathbb{E}[X^3],a+6,mathbb{E}[X^2],a^2-4,mathbb{E}[X],a^3+a^4,.$$
Thus,
$$f'(a)=-4,mathbb{E}[X^3]+12,mathbb{E}[X^2],a-12,mathbb{E}[X],a^2+4,a^3,.$$
Therefore, if $a=mathbb{E}[X]$ minimizes $f$, then $f'big(mathbb{E}[X]big)=0$, and so
$$-mathbb{E}[X^3]+3,mathbb{E}[X^2],mathbb{E}[X]-2,big(mathbb{E}[X]big)^3=0,.tag{*}$$
Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $mathbb{E}[X]=0$ but $mathbb{E}[X^3]neq 0$.
Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$mathbb{P}[X=-1]=frac{2}{3}text{ and }mathbb{P}[X=2]=frac13,.$$ Then, $$mathbb{E}[X]=0text{ whereas }mathbb{E}[X^3]=2neq 0,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18left(sqrt[3]{2}-1right)neq 0=mathbb{E}[X],.$$
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
$endgroup$
Expanding $mathbb{E}left[(X-a)^4right]$ yields
$$f(a):=mathbb{E}left[(X-a)^4right]=mathbb{E}[X^4]-4,mathbb{E}[X^3],a+6,mathbb{E}[X^2],a^2-4,mathbb{E}[X],a^3+a^4,.$$
Thus,
$$f'(a)=-4,mathbb{E}[X^3]+12,mathbb{E}[X^2],a-12,mathbb{E}[X],a^2+4,a^3,.$$
Therefore, if $a=mathbb{E}[X]$ minimizes $f$, then $f'big(mathbb{E}[X]big)=0$, and so
$$-mathbb{E}[X^3]+3,mathbb{E}[X^2],mathbb{E}[X]-2,big(mathbb{E}[X]big)^3=0,.tag{*}$$
Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $mathbb{E}[X]=0$ but $mathbb{E}[X^3]neq 0$.
Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$mathbb{P}[X=-1]=frac{2}{3}text{ and }mathbb{P}[X=2]=frac13,.$$ Then, $$mathbb{E}[X]=0text{ whereas }mathbb{E}[X^3]=2neq 0,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18left(sqrt[3]{2}-1right)neq 0=mathbb{E}[X],.$$
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
edited Dec 14 '18 at 21:47
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
answered Dec 14 '18 at 21:14
BatominovskiBatominovski
33.1k33293
33.1k33293
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
add a comment |
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
1
1
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
You can take the derivative first and get $4E(X-a)^3)$, then expand out.
$endgroup$
– Acccumulation
Dec 14 '18 at 21:36
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
$begingroup$
Without loss of generality we can reduce the OP's conjecture to $Bbb E[(X-a)^4]$ being minimised at $a=0$ if $Bbb E[X]=0$, so (*) becomes $Bbb E[X^3]=0$ (provided $X$ has finite variance), which won't hold for a skewed distribution.
$endgroup$
– J.G.
Dec 14 '18 at 22:22
add a comment |
$begingroup$
For a counter example take a look at a random variable with $Xin{1,-2}$ with probabilities $$P(X=+1) =frac{2}{3}, qquad P(X=-2) = frac{1}{3},.$$
Now, we can see that
$$E(X) =frac{2}{3} cdot 1 + frac{1}{3}cdot(-2) = 0,.$$
However,
$$E((X-a)^4) = frac{2}{3} cdot (1-a)^4 + frac{1}{3}cdot(-2-a)^4,.$$
We have
$$ frac{81}{16}= E((X+tfrac12)^4) < E(X^4) = 6$$
providing a counterexample.
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
$endgroup$
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
add a comment |
$begingroup$
For a counter example take a look at a random variable with $Xin{1,-2}$ with probabilities $$P(X=+1) =frac{2}{3}, qquad P(X=-2) = frac{1}{3},.$$
Now, we can see that
$$E(X) =frac{2}{3} cdot 1 + frac{1}{3}cdot(-2) = 0,.$$
However,
$$E((X-a)^4) = frac{2}{3} cdot (1-a)^4 + frac{1}{3}cdot(-2-a)^4,.$$
We have
$$ frac{81}{16}= E((X+tfrac12)^4) < E(X^4) = 6$$
providing a counterexample.
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
$endgroup$
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
add a comment |
$begingroup$
For a counter example take a look at a random variable with $Xin{1,-2}$ with probabilities $$P(X=+1) =frac{2}{3}, qquad P(X=-2) = frac{1}{3},.$$
Now, we can see that
$$E(X) =frac{2}{3} cdot 1 + frac{1}{3}cdot(-2) = 0,.$$
However,
$$E((X-a)^4) = frac{2}{3} cdot (1-a)^4 + frac{1}{3}cdot(-2-a)^4,.$$
We have
$$ frac{81}{16}= E((X+tfrac12)^4) < E(X^4) = 6$$
providing a counterexample.
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
$endgroup$
For a counter example take a look at a random variable with $Xin{1,-2}$ with probabilities $$P(X=+1) =frac{2}{3}, qquad P(X=-2) = frac{1}{3},.$$
Now, we can see that
$$E(X) =frac{2}{3} cdot 1 + frac{1}{3}cdot(-2) = 0,.$$
However,
$$E((X-a)^4) = frac{2}{3} cdot (1-a)^4 + frac{1}{3}cdot(-2-a)^4,.$$
We have
$$ frac{81}{16}= E((X+tfrac12)^4) < E(X^4) = 6$$
providing a counterexample.
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
edited Dec 14 '18 at 21:20
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
answered Dec 14 '18 at 21:19
FabianFabian
19.8k3774
19.8k3774
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
add a comment |
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
$begingroup$
We both came up with very similar counterexamples.
$endgroup$
– Batominovski
Dec 14 '18 at 21:19
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039880%2fprove-disprove-that-e-bigx%25e2%2588%2592a4-big-is-minimized-when-a-ex%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Can you express $E((X-a)^4)$ as a function of $E((X-a)^2)$? Hint: variance.
$endgroup$
– Frank Vel
Dec 14 '18 at 21:08