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The law of the unconscious statistician computes the expected value of a measurable function of a random variable directly from the original probability distribution. If $g(X)$ is an integrable random variable, thenUse ${\sum }_{x}g(x)p_X(x)$ for a probability mass function, or ${\int }_{\mathbb{R}}g(x)f_X(x)dx$ for a probability density function. There is no need to derive the probability distribution of $g(X)$ first. The identity is the integration rule for a pushforward measure, and also holds with an infinite value when $g$ is nonnegative. ^[ref].

For $X$ taking $-1,2,8$ with probabilities $1/2,3/10,1/5$, choosing $g(x)=x^2$ gives $\mathbb{E}[X^2]=\frac{1}{2}+4\frac{3}{10}+64\frac{1}{5}=14.5$. Its variance is therefore $14.5-1.7^2=11.61$. Squaring the expected value would instead give $2.89$, because squaring does not commute with averaging.

For integrable random variables $X_1,\dots ,X_n$ on the same probability space and fixed real numbers $a_1,\dots ,a_n$, linearity of expectation statesThe random variables can be dependent random variables. The formula follows by the linearity of integration; a finite linear combination of integrable random variables remains integrable. For example, if $X$ takes $-1,2,8$ with probabilities $1/2,3/10,1/5$, then $X$ and $2X$ are dependent random variables, yet $\mathbb{E}[X+2X]=3\mathbb{E}[X]=5.1$.

An especially useful application counts how many events occur. Put $I_i=1_{A_i}$, an indicator random variable, and $N={\sum }_i{I}_i$. Since $\mathbb{E}[I_i]=\mathbb{P}(A_i)$, we get $\mathbb{E}[N]={\sum }_i\mathbb{P}(A_i)$. There is no need to compute the whole probability distribution of $N$, or to assume independence. For a uniform random permutation of $n\ge 1$ objects, each object stays fixed with probability $1/n$, so the fixed point count of a uniform random permutation has expected value one for every $n$. ^[ref].

Linearity of expectation concerns addition and multiplication by fixed scalars. Products instead satisfy $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ under the hypotheses of expectation of a product of independent random variables. For the example above, $\mathbb{E}[X^2]=14.5$, whereas $(\mathbb{E}[X]{)}^2=2.89$. Applying a nonlinear function before averaging can therefore change the answer substantially; use the law of the unconscious statistician to compute that average.

The law of the unconscious statistician gives the improper integralIn particular $\mathbb{E}(X)=1/2$ and $\mathbb{E}(X^2)=1/3$, soFor $p=-1$, ${\int }_{\epsilon }^1{x}^{-1}dx=-\log \epsilon \to +\infty$; for $p<-1$, the power integral also tends to $+\infty$. Thus $\mathbb{E}(X^p)=+\infty$ for $p\le -1$, so there is no finite expected value. Negative powers need only be defined almost surely on $X>0$; the null event $X=0$ has no effect on this conclusion.

The law of the unconscious statistician and the finite variance giveThus size bias increases the mean by ${\sigma }^2/\mu$, in accord with preferentially selecting larger values.