Expected Value

A random variable can take on a large, often infinite, number of values. We are interested in substituting, in place of the possible values of the random variable, a representative value that takes into account the large or small probabilities associated with the RV. This value is called the Expected Value or Expectation (in italiano valore atteso o speranza matematica) and its symbol is $E[\cdot]$. In the discrete case we can define such a value as an average of the observations weighted with the respective probabilities, while in the continuous case we need to substitute sums with integrals:

• Let $X$ be a discrete random variable and let $P(X=x_k) = P_k$. Then
  $$E[X] = \sum_{k=1}^\infty x_k p_k$$
  If $\phi(x)$ is a continuous function then:
  $$E[\phi(X)] = \sum_{k=1}^\infty \phi(x_k) p_k$$

• Let $X$ be a continous random variable with continous density $f_X$. Then
  $$E[X] = \int_{-\infty}^{+\infty} t f_X(t) dt$$
  If $\phi(x)$ is a continuous function then:
  $$E[\phi(X)] = \int_{-\infty}^{+\infty} \phi(t) f_X(t) dt$$

It can be easily verified that

$$E[\alpha X + \beta Y] = \alpha EX + \beta EY$$

Examples:

• if $X\sim N(0,1)$, then
  

  $$E[X] = \int_{-\infty}^{+\infty} t \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}}dt$$

  $$= \left. \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\right\vert _{-\infty}^{+\infty}$$

  $$=$$ 0

  
  
• if $X\sim N(\mu,\sigma^2)$, then
  

  $$E[X] = \int_{-\infty}^{+\infty} \frac{t}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}(\frac{t-mu}{\sigma})^2}dt$$

  $$(z=\frac{t-\mu}{\sigma} \Rightarrow dz = \frac{dt}{\sigma}$$

  $$= \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}} (\sigma z + \mu) e^{-\frac{z^2}{2}}dz$$

  $$= \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} z e^{-\frac{z^2}{2}} dz + \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-\frac{z^2}{2}} dz$$

  $$= \mu$$

  
  
• if $X\sim Exp(\lambda)$
  

  $$E[X] = \int_{0}^{+\infty} t e^{-\lambda t}dt$$

  $$= \left. \lambda e^{-\lambda t}\right\vert _0^{+\infty} - \int_{0}^{+\infty} e^{-\lambda t}dt$$ (p.p.)

  $$= \frac{1}{\lambda}$$

  
  

We are also interested in measuring the discrepancies with respect to the expected value. This is accomplished by the ``variance'':

   var$$[X] = E[(X - E X)^2] = E[X^2] + E[X]^2-2E[X]E[X] = E[X^2]-E[X]^2$$

Examples:

• if $X\sim N(0,1)$, then
  

  $$E[X^2] = \int_{-\infty}^{+\infty} t^2 \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}}dt$$

  $$= \frac{1}{\sqrt{2\pi}} \left[\left. -te^{-\frac{t^2}{2}}\right\vert _{-\infty}^{+\infty} + \int_{-\infty}{^{+\infty}e;-\frac{t^2}{2}}dt\right]$$ (p.p.)

  $$= 1$$

  
  and since $E[X]=0$ we have:
     var$$[X] = 1$$

• if $Y\sim N(\mu,\sigma^2)$, then we can write $Y=\sigma X + \mu$ with $X\sim N(0,1)$. It follows:
  $$E[Y] = \mu ; E[X^2]=1$$
     var$$[Y] = E[(y-\mu)^2]=E[(\sigma X + \mu - \mu)^2] = \sigma^2E[X^2] = \sigma^2$$

• if $X\sim U[0,1]$, we say that a random variable $X$ has uniform distribution in $[0,1]$. Its density is shown in Figure 1.6, and its cumulative density is as represented in figure 1.7. A simple calculation shows that $E[X]=1/2$ and var$[X]=1/12$.

  Figure 1.6: Uniform density in $[0,1]$.

  

  Figure 1.7: Uniform CDF in $[0,1]$.