Expectation of a continuous random variable with probability one at a given point.

Question

$$f_{X}(x) = \begin{cases}\frac{x+1}{a+1} & -1 \leq x < a \\ 1 & x=a \\ \frac{x-1}{a-1} & a < x \leq 1 \\ 0 & \text{otherwise}\end{cases}$$

Calculate $E(X)$

I know typically this is a very easy solution (sum the three integrals, integrating between the limits given in the PDF) but in this case, the $P_X$(x)=1 when x=a is throwing me a little. Is that integral =1, a, $a^2$/2 or 0? I suppose the question then is really what should my limits on that integration be? Or even, can I even integrate that, bearing in mind its a point function?

Thanks!

P.S, I apologis for my formatting; please feel free to edit.

Answer 1

The probabiity density function at $x=a$ takes the value $1$ does not imply that the probability at $x=a$ is $1$.
Because the probability at $x=a$ is given by $P(x=a)=P(a\le x \le a)=\int^{a}_{a}f(x) dx = 0$
So $$E(X)=\int^{1}_{-1} xf(x) dx=\int^{a}_{-1}\frac{x^2+x}{a+1} dx +\int^{a}_{a}x dx+\int^{1}_{a} \frac{x^2-x}{a-1} dx$$

Now you see what you need to do.

Answer 2

As others have said, the value of a function at a single point is irrelevant to an integral of that function. I will add, however, that before you can determine the expectation, you will have to determine the correct value of "a"- that value such that the "total probability", the integral of that probability density function from -1 to 1, is 1.

I spoke to soon- that integral is 1 for all a!