একটি সীমানা র্যান্ডম ভেরিয়েবলের বৈকল্পিক

Question

মনে করুন যে এলোমেলো ভেরিয়েবলের একটি নিম্ন এবং উপরের বাউন্ড [0,1] রয়েছে। এ জাতীয় চলকের বৈকল্পিক কীভাবে গণনা করা যায়?

Comments

আনবাউন্ডেড ভেরিয়েবলের মতো একই পদ্ধতি - সংহতকরণ বা সংমিশ্রণের সীমা যথাযথভাবে সেট করে।

— Scortchi - পুনর্বহাল মনিকা

যেমন @ স্কার্টচি বলেছেন। তবে আমি কৌতূহল করছি কেন আপনি ভাবেন যে এটি অন্যরকম হতে পারে?

— পিটার ফ্লুম - মনিকা পুনরায়

আপনি যদি পরিবর্তনশীল সম্পর্কে কিছু না জানেন (তবে সেক্ষেত্রে ভিন্নতার উপরের একটি উপরের আবদ্ধটি সীমানার অস্তিত্ব থেকে গণনা করা যেতে পারে), কেন এটি বেঁধে রাখা সত্যটি গণনায় আসবে?

— গ্লেন_বি -রিনস্টেট মনিকা

একটি দরকারী উপরের একটি এলোপাতাড়ি ভেরিয়েবলের যে মান লাগে ভ্যারিয়েন্স উপর আবদ্ধ $[a,b]$ সম্ভাব্যতা সঙ্গে $1$ হয় $(b-a)^2/4$ এবং একটি বিযুক্ত দৈব চলক যে মান লাগে করে এটা করা যায় $a$ এবং $b$ সমান সঙ্গে সম্ভাবনা $\frac{1}{2}$ । আরেকটি বিষয় মনে রাখবেন যে ভেরিয়েন্সটির উপস্থিতি গ্যারান্টিযুক্ত যেখানে একটি আনবাউন্ডেড এলোমেলো ভেরিয়েবলের কোনও ভেরিয়েন্স নাও থাকতে পারে (কিছু, যেমন কচির এলোমেলো ভেরিয়েবলের অর্থও না থাকে)।

— দিলিপ সরোতে

সেখানে হয় একটি বিযুক্ত দৈব চলক যার ভ্যারিয়েন্স সমান $\frac{(b-a)^2}{4}$ ঠিক:একটি এলোপাতাড়ি ভেরিয়েবলের যে মান লাগে$a$এবং$b$সমান সম্ভাবনা থাকে$\frac{1}{2}$ । সুতরাং, কমপক্ষে আমরা জানি যে বৈকল্পিকের উপর একটি সার্বজনীন উপরের আবদ্ধ(বি-এ)2 এরচেয়ে ছোট হতে পারে না । $\frac{(b-a)^2}{4}$

— দিলিপ সরোতে

Answer 1

আপনি নীচে Popoviciu এর বৈষম্য প্রমাণ করতে পারেন। চিহ্নিতকরণটি $m=\inf X$ এবং $M=\sup X$ । G ( t ) = E দ্বারা একটি ফাংশন $g$ সংজ্ঞায়িত করুন [ ( এক্স - টি ) 2 ]

$$g(t)=\mathbb{E}\left[\left(X-t\right)^2\right] \, .$$ ব্যুৎপন্ন কম্পিউটিং$g'$ , এবং সমাধানে $$g'(t) = -2\mathbb{E}[X] +2t=0 \, ,$$ আমরা যে$g$ তার ন্যূনতম অর্জন$t=\mathbb{E}[X]$ (নোট যে$g''>0$ )।

এখন, ফাংশনের মান বিবেচনা $g$ বিশেষ সময়ে $t=\frac{M+m}{2}$ । এটি অবশ্যই হওয়া উচিত যে

$$\mathbb{Var}[X]=g(\mathbb{E}[X])\leq g\left(\frac{M+m}{2}\right) \, .$$ তবে $$g\left(\frac{M+m}{2}\right) = \mathbb{E}\left[\left(X - \frac{M+m}{2}\right)^2 \right] = \frac{1}{4}\mathbb{E}\left[\left((X-m) + (X-M)\right)^2 \right] \, .$$ যেহেতু$X-m\geq 0$ এবং$X-M\leq 0$ , আমরা $$\left((X-m)+(X-M)\right)^2\leq\left((X-m)-(X-M)\right)^2=\left(M-m\right)^2 \, ,$$ যে 1 বোঝা $$\frac{1}{4}\mathbb{E}\left[\left((X-m) + (X-M)\right)^2 \right] \leq \frac{1}{4}\mathbb{E}\left[\left((X-m) - (X-M)\right)^2 \right] = \frac{(M-m)^2}{4} \, .$$ অতএব, আমরা Popoviciu এর বৈষম্য প্রমাণিত $$\mathbb{Var}[X]\leq \frac{(M-m)^2}{4} \, .$$

Comments

Nice approach: it's good to see rigorous demonstrations of these kinds of things.

— whuber

+1 Nice! I learned statistics long before computers were in vogue, and one idea that was drilled into us was that $$E[(X-t)^2] = E[((X-\mu)-(t-\mu))^2] = E[(X-\mu)^2]+(t-\mu)^2$$ which allowed for the computation of variance by finding the sum of the squares of the deviations from any convenient point $t$ and then adjusting for the bias. Here of course, this identity gives a simple proof of the result that $g(t)$ has minimum value at $t=\mu$ without the necessity of derivatives etc.

— Dilip Sarwate

Answer 2

Let $F$ be a distribution on $[0,1]$. We will show that if the variance of $F$ is maximal, then $F$ can have no support in the interior, from which it follows that $F$ is Bernoulli and the rest is trivial.

As a matter of notation, let $\mu_k = \int_0^1 x^k dF(x)$ be the $k$th raw moment of $F$ (and, as usual, we write $\mu = \mu_1$ and $\sigma^2 = \mu_2 - \mu^2$ for the variance).

We know $F$ does not have all its support at one point (the variance is minimal in that case). Among other things, this implies $\mu$ lies strictly between $0$ and $1$. In order to argue by contradiction, suppose there is some measurable subset $I$ in the interior $(0,1)$ for which $F(I)\gt 0$. Without any loss of generality we may assume (by changing $X$ to $1-X$ if need be) that $F(J = I \cap (0, \mu]) \gt 0$: in other words, $J$ is obtained by cutting off any part of $I$ above the mean and $J$ has positive probability.

Let us alter $F$ to $F'$ by taking all the probability out of $J$ and placing it at $0$. In so doing, $\mu_k$ changes to

$$\mu'_k = \mu_k - \int_J x^k dF(x).$$

As a matter of notation, let us write $[g(x)] = \int_J g(x) dF(x)$ for such integrals, whence

$$\mu'_2 = \mu_2 - [x^2], \quad \mu' = \mu - [x].$$

Calculate

$$\sigma'^2 = \mu'_2 - \mu'^2 = \mu_2 - [x^2] - (\mu - [x])^2 = \sigma^2 + \left((\mu[x] - [x^2]) + (\mu[x] - [x]^2)\right).$$

The second term on the right, $(\mu[x] - [x]^2)$, is non-negative because $\mu \ge x$ everywhere on $J$. The first term on the right can be rewritten

$$\mu[x] - [x^2] = \mu(1 - [1]) + ([\mu][x] - [x^2]).$$

The first term on the right is strictly positive because (a) $\mu \gt 0$ and (b) $[1] = F(J) \lt 1$ because we assumed $F$ is not concentrated at a point. The second term is non-negative because it can be rewritten as $[(\mu-x)(x)]$ and this integrand is nonnegative from the assumptions $\mu \ge x$ on $J$ and $0 \le x \le 1$. It follows that $\sigma'^2 - \sigma^2 \gt 0$.

We have just shown that under our assumptions, changing $F$ to $F'$ strictly increases its variance. The only way this cannot happen, then, is when all the probability of $F'$ is concentrated at the endpoints $0$ and $1$, with (say) values $1-p$ and $p$, respectively. Its variance is easily calculated to equal $p(1-p)$ which is maximal when $p=1/2$ and equals $1/4$ there.

Now when $F$ is a distribution on $[a,b]$, we recenter and rescale it to a distribution on $[0,1]$. The recentering does not change the variance whereas the rescaling divides it by $(b-a)^2$. Thus an $F$ with maximal variance on $[a,b]$ corresponds to the distribution with maximal variance on $[0,1]$: it therefore is a Bernoulli$(1/2)$ distribution rescaled and translated to $[a,b]$ having variance $(b-a)^2/4$, QED.

Comments

Interesting, whuber. I didn't know this proof.

— Zen

@Zen It's by no means as elegant as yours. I offered it because I have found myself over the years thinking in this way when confronted with much more complicated distributional inequalities: I ask how the probability can be shifted around in order to make the inequality more extreme. As an intuitive heuristic it's useful. By using approaches like the one laid out here, I suspect a general theory for proving a large class of such inequalities could be derived, with a kind of hybrid flavor of the Calculus of Variations and (finite dimensional) Lagrange multiplier techniques.

— whuber

Perfect: your answer is important because it describes a more general technique that can be used to handle many other cases.

— Zen

@whuber said - "I ask how the probability can be shifted around in order to make the inequality more extreme." -- this seems to be the natural way to think about such problems.

— Glen_b -Reinstate Monica

There appear to be a few mistakes in the derivation. It should be $$\mu[x] - [x^2] = \mu(1 - [1])[x] + ([\mu][x] - [x^2]).$$ Also, $[(\mu-x)(x)]$ does not equal $[\mu][x] - [x^2]$ since $[\mu][x]$ is not the same as $\mu[x]$

— Leo

Answer 3

If the random variable is restricted to $[a,b]$ and we know the mean $\mu=E[X]$, the variance is bounded by $(b-\mu)(\mu-a)$.

Let us first consider the case $a=0, b=1$. Note that for all $x\in [0,1]$, $x^2\leq x$, wherefore also $E[X^2]\leq E[X]$. Using this result,

$$\begin{equation} \sigma^2 = E[X^2] - (E[X]^2) = E[X^2] - \mu^2 \leq \mu - \mu^2 = \mu(1-\mu). \end{equation}$$

To generalize to intervals $[a,b]$ with $b>a$, consider $Y$ restricted to $[a,b]$. Define $X=\frac{Y-a}{b-a}$, which is restricted in $[0,1]$. Equivalently, $Y = (b-a)X + a$, and thus

$$\begin{equation} Var[Y] = (b-a)^2Var[X] \leq (b-a)^2\mu_X (1-\mu_X). \end{equation}$$ where the inequality is based on the first result. Now, by substituting $\mu_X = \frac{\mu_Y - a}{b-a}$, the bound equals $$\begin{equation} (b-a)^2\, \frac{\mu_Y - a}{b-a}\,\left(1- \frac{\mu_Y - a}{b-a}\right) = (b-a)^2 \frac{\mu_Y -a}{b-a}\,\frac{b - \mu_Y}{b-a} = (\mu_Y - a)(b- \mu_Y), \end{equation}$$ which is the desired result.

Answer 4

At @user603's request....

A useful upper bound on the variance $\sigma^2$ of a random variable that takes on values in $[a,b]$ with probability $1$ is $\sigma^2 \leq \frac{(b−a)^2}{4}$. A proof for the special case $a=0, b=1$ (which is what the OP asked about) can be found here on math.SE, and it is easily adapted to the more general case. As noted in my comment above and also in the answer referenced herein, a discrete random variable that takes on values $a$ and $b$ with equal probability $\frac{1}{2}$ has variance $\frac{(b−a)^2}{4}$ and thus no tighter general bound can be found.

Another point to keep in mind is that a bounded random variable has finite variance, whereas for an unbounded random variable, the variance might not be finite, and in some cases might not even be definable. For example, the mean cannot be defined for Cauchy random variables, and so one cannot define the variance (as the expectation of the squared deviation from the mean).

Comments

this is a special case of @Juho's answer

— Aksakal

It was just a comment, but I could also add that this answer does not answer the question asked.

— Aksakal

@Aksakal So??? Juho was answering a slightly different and much more recently asked question. This new question has been merged with the one you see above, which I answered ten months ago.

— Dilip Sarwate

Answer 5

are you sure that this is true in general - for continuous as well as discrete distributions? Can you provide a link to the other pages? For a general distibution on $[a,b]$ it is trivial to show that

$$Var(X) = E[(X-E[X])^2] \le E[(b-a)^2] = (b-a)^2.$$ I can imagine that sharper inequalities exist ... Do you need the factor $1/4$ for your result?

On the other hand one can find it with the factor $1/4$ under the name Popoviciu's_inequality on wikipedia.

This article looks better than the wikipedia article ...

For a uniform distribution it holds that

$$Var(X) = \frac{(b-a)^2}{12}.$$

Comments

This page states the result with the start of a proof that gets a bit too involved for me as it seems to require an understanding of the "Fundamental Theorem of Linear Programming". sci.tech-archive.net/Archive/sci.math/2008-06/msg01239.html

— Adam Russell

Thank you for putting a name to this! "Popoviciu's Inequality" is just what I needed.

— Adam Russell

This answer makes some incorrect suggestions: $1/4$ is indeed right. The reference to Popoviciu's inequality will work, but strictly speaking it applies only to distributions with finite support (in particular, that includes no continuous distributions). A limiting argument would do the trick, but something extra is needed here.

— whuber

A continuous distribution can approach a discrete one (in cdf terms) arbitrarily closely (e.g. construct a continuous density from a given discrete one by placing a little Beta(4,4)-shaped kernel centered at each mass point - of the appropriate area - and let the standard deviation of each such kernel shrink toward zero while keeping its area constant). Such discrete bounds as discussed here will thereby also act as bounds on continuous distributions. I expect you're thinking about continuous unimodal distributions... which indeed have different upper bounds.

— Glen_b -Reinstate Monica

Well ... my answer was the least helpful but I would leave it here due to the nice comments. Cheers,R

— Ric