Variance d'une variable aléatoire bornée

Supposons qu'une variable aléatoire ait une borne inférieure et une borne supérieure [0,1]. Comment calculer la variance d'une telle variable?

Vous pouvez prouver l'inégalité de Popoviciu comme suit. Utiliser la notation $m=\inf X$ et $M=\sup X$ . Définissez une fonction $g$ par

Maintenant, considérons la valeur de la fonction $g$ au point spécial $t=\frac{M+m}{2}$ . Ce doit être le cas que

Soit F une distribution sur [ 0 , 1 ] . Nous montrerons que si la variance de F est maximale, alors F ne peut avoir aucun support à l'intérieur, d'où il résulte que F est Bernoulli et le reste est trivial.$F$$[0,1]$$F$$F$$F$

En termes de notation, soit μ k = ∫ 1 0 x k d F ( x ) le k ème moment brut de F (et, comme d'habitude, on écrit μ = μ 1 et σ 2 = μ 2 - μ 2 pour la variance).$\mu_k = \int_0^1 x^k dF(x)$$k$$F$$\mu = \mu_1$$\sigma^2 = \mu_2 - \mu^2$

Nous savons que F n'a pas tout son support à un moment donné (la variance est minime dans ce cas). Cela implique entre autres que μ se situe strictement entre 0 et 1 . Pour argumenter par contradiction, supposons qu'il existe un sous-ensemble mesurable I à l'intérieur ( 0 , 1 ) pour lequel F ( I ) > 0 . Sans aucune perte de généralité, nous pouvons supposer (en changeant X en 1 - X si besoin est) que F ( J = $F$$\mu$$0$$1$$I$$(0,1)$$F(I)\gt 0$$X$$1-X$∩ ( 0 , μ ] ) > 0 : en d'autres termes, J est obtenu en coupant toute partie de I au-dessus de la moyenne et J a une probabilité positive.$F(J = I \cap (0, \mu]) \gt 0$$J$$I$$J$

Modifions F en F ′ en retirant toute probabilité de J et en la plaçant à 0 . $F$$F'$$J$$0$ Ce faisant, μ k devient$\mu_k$

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

Calculate

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

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.

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,

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

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).

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

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