有界随机变量的方差

假设随机变量具有上限和下限[0,1]。如何计算这样一个变量的方差？

您可以证明波波维丘不等式如下。使用符号$m=\inf X$和$M=\sup X$。通过 g （t ）= E [ （X - t ） 2 ]定义函数$g$

现在，考虑在特殊点t = M + m处的函数$g$的值$t=\frac{M+m}{2}$。它必须是这样的情况

让˚F是对分布[ 0 ，1 ]。我们将证明，如果方差˚F是最大的，那么˚F可以有没有在内部支持，从中可以得出˚F是伯努利，其余是微不足道的。$F$$[0,1]$$F$$F$$F$

作为符号的问题，让μ ķ = ∫ 1 0 X ķ d ˚F （X ）是ķ个的生力矩˚F（和往常一样，我们写μ = μ 1和σ 2 = μ 2 - μ 2为差异）。$\mu_k = \int_0^1 x^k dF(x)$$k$$F$$\mu = \mu_1$$\sigma^2 = \mu_2 - \mu^2$

我们知道F在某一时刻并没有获得所有支持（在这种情况下，方差很小）。除其他外，这意味着μ严格位于0到1之间。为了通过矛盾争辩，假设有一些可测量的子集我在内部（0 ，1 ）为其˚F （我）> 0。不失一般性的任何损失我们可以假定（通过改变X到1 - X如果需要的话），该˚F （Ĵ = $F$$\mu$$0$$1$$I$$(0,1)$$F(I)\gt 0$$X$$1-X$∩ （0 ，μ ] ）> 0：换句话说，通过切除均值以上的 I的任何部分来获得 J，并且 J具有正概率。$F(J = I \cap (0, \mu]) \gt 0$$J$$I$$J$

让我们改变˚F到˚F '采取一切概率出Ĵ并将其放置在0。$F$$F'$$J$$0$ 在这样做时，μ ķ变为$\mu_k$

作为符号的问题，让我们写出[ 克（X ）] = ∫ Ĵ克（X ）d ˚F （X ）这样的积分，从那里$[g(x)] = \int_J g(x) dF(x)$

计算

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