Expectation of a function of a continuous random variable

[kingwinner]

If W=g(X) is a function of continuous random variable X, then E(W)=E[g(X)]=
∞
∫g(x) [f_X(x)] dx
-∞
============================
Even though X is continuous, g(X) might not be continuous.

If W happens to be a discrete random variable, does the above still hold? Do we still integrate ∫ (instead of sum ∑)?

Does it matter whether W itself is discrete or continuous?

Thanks!

 

[jambaugh]

If the variable is discrete you can still express in terms of integrals using dirac delta functions though it is simpler to replace the integral with a sum over values.

 

[jason1995]

If W=g(X) is a function of continuous random variable X, then E(W)=E[g(X)]=
∞
∫g(x) [f_X(x)] dx
-∞
============================
Even though X is continuous, g(X) might not be continuous.

If W happens to be a discrete random variable, does the above still hold?

Yes. Consider the simplest case, where g(x) = 1 if x is in [a,b] and 0 otherwise. Then

$$\begin{align*}<br /> E[W] &= 1\cdot P(W = 1) + 0\cdot P(W = 0)\\<br /> &= P(g(X) = 1)\\<br /> &= P(X\in [a,b])\\<br /> &= \int_a^b f_X(x)\,dx.<br /> \end{align*}$$

On the other hand, the formula also gives

$$\int_{-\infty}^\infty g(x)f_X(x)\,dx = \int_a^b f_X(x)\,dx.$$

 

[gkannan16]

Go for summations...that will solve ur problem and rest all may complicate