# Cumulative distribution function and expected value

I've got cumulative distribution function given: $F_X(t) = 0$ for $t<0$, $\frac{1}{3}$ for $t=0$ , $\frac{1}{3} + \frac{t}{90}$ for $t\in (0,60)$ and $1$ for $t \ge 60$. I am to find expexted value ($EX$). So, what to do? I wanted to find a density function, but I can't, because cumulative distribution function is not continuous. So what am I supposed to do here?

You do not need to find a density function. If a random variable takes on positive values only (or is positive with probability $1$), then its expected value is $$E[X] = \int_0^\infty [1-F_X(t)]\,\mathrm dt.$$

Since you are given $F_X(t)$, just compute the value of the above integral. In this instance, since $F(t)$ increases linearly from $\frac{1}{3}$ at $t=0$ to $1$ at $t=60$, $[1-F_X(t)]$ is decreasing linearly from $\frac{2}{3}$ at $t=0$ to $0$ at $t=60$, that is, you are finding the area of a (right) triangle which you can do without necessarily formally integrating.

You can certainly write the density function for each interval of interest, and make sure you account for the normalization constant (such that your pdf integrates to 1). Then, use the definition of $E(X)$.

You can find a pdf. It's given by $$p_X(t) = \frac{1}{3}\delta(t) + \frac{1}{90},\quad 0<t<60$$ (zero otherwise). So $E\{X\}$ is given by

$$E\{X\} = \int_{0}^{60}t\left ( \frac{1}{3}\delta(t) + \frac{1}{90}\right )\;dt$$

So we get

$$E\{X\} = \frac{1}{3}\int_{0}^{60}t\delta(t)\;dt + \frac{1}{90}\int_{0}^{60}t\;dt = \frac{1}{90}\frac{60^2}{2} = 20$$

• And what is $\delta(t)$? – Anne Apr 29 '13 at 21:11
• It's the Dirac delta function: $\int_{-\infty}^{\infty}\delta(t)dt = 1$ and $\int_{-\infty}^{\infty}f(t)\delta(t)dt=f(0)$. – Matt L. Apr 30 '13 at 5:55

Another way of looking at it is using the Riemann-Stieltjes integral. In this case, we have

$$E[X] = \int_{-\infty}^\infty t\ dF_X(t).$$

The advantage here is that $F_X(t)$ satisfies all the conditions necessary for the Riemann-Stieltjes integral to exist: it is monotonic, and even though it has discontinuities, the function $f(t) = t$ is everywhere continuous.

By using properties of the Riemann-Stieljes integral, we can easily compute your solution simply by separating the integral into piecewise components corresponding to the breakpoints of your piecewise linear CDF. Since $F_X(t)$ is constant above $t=60$ and below $t=0$, it is continuous and we can simply note that $dF_X(t) = 0$ in these regions.

This gets you exactly to Matt L.'s solution, but maybe it seems a little less like magic.