Varianz einer begrenzten Zufallsvariablen

Angenommen, eine Zufallsvariable hat eine untere und eine obere Schranke [0,1]. Wie berechnet man die Varianz einer solchen Variablen?

Sie können Popovicius Ungleichung wie folgt beweisen. Verwenden die Notation $m=\inf X$ und $M=\sup X$ . Definiere eine Funktion $g$ durch

Betrachten Sie nun den Wert der Funktion $g$ am Sonderpunkt $t=\frac{M+m}{2}$ . Es muss der Fall seindass

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

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