Law of the Unconscious Statistician

Topics: Probability - Expected Value

(theorem)

Let $X$ be a random variable and let $g:\mathbb{R}\to \mathbb{R}$ be a function such that $Y=g(x)$ is also a random variable.

If we wanted to calculate $\mathbb{E}[Y]$, the expected value of $Y$, we would have to first find $f_Y$ (its density function). There is, however, a simpler way to calculate it, called the law of the unconscious statistician.

Example

Let $X$ be a discrete random variable where $\mathbb{E}[X]=0$ whose density function is defined as:

$$f_X(x)=\{\begin{array}{ll}\frac{1}{9},& \text{ if }x\in \{-4,-3,-2,-1,0,1,2,3,4\}\\ 0,& \text{ otherwise}\end{array}$$

Let $Z=(X-\mathbb{E}[X]{)}^2=X^2$. What’s $\mathbb{E}[Z]$?

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Normal way:

We can calculate $\mathbb{E}[Z]$ in the normal way. We first determine its density function:

$$f_Z(z)=\{\begin{array}{l}\frac{1}{9},\text{ if }z=0\\ \frac{2}{9},\text{ if }z=1,4,9,16\\ 0,\text{ otherwise}\end{array}$$

Then we use the definition of an expected value:

$$\begin{array}{rl}\mathbb{E}[Z]& =\sum _{x=0}^{16}z\mathbb{P}(Z=z)\\ & =(0)\frac{1}{9}+(1)\frac{2}{9}+(4)\frac{2}{9}+(9)\frac{2}{9}+(16)\frac{2}{9}\\ & =\frac{60}{9}\end{array}$$

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With the law of the unconscious statistician:

However, it is much easier if we just use the law of the unconscious statistician. Since $X$ is a discrete random variable, then:

$$\begin{array}{rl}\mathbb{E}[Z]=\mathbb{E}[X^2]& =\sum _{x=-4}^4{x}^2\mathbb{P}(X=x)\\ & =(-4{)}^2\frac{1}{9}+(-3{)}^2\frac{1}{9}+(-2{)}^2\frac{1}{9}+(-1{)}^2\frac{1}{9}\\ & +0^2\frac{1}{9}+1^2\frac{1}{9}+2^2\frac{1}{9}+3^2\frac{1}{9}+4^2\frac{1}{9}\\ & =\frac{60}{9}\end{array}$$

Discrete Case §

If $X$ is a discrete random variable, then:

$$\mathbb{E}[Y]=\mathbb{E}[g(x)]=\sum _{x\in R_X}g(x)\mathbb{P}(X=x)$$

Continuous Case §

If $X$ is an absolutely continuous random variable, then:

$$\mathbb{E}[Y]=\mathbb{E}[g(x)]={\int }_{\mathbb{R}}g(x)f_X(x)dx$$