The expected value of a function

Question

What is the expected value of $.1\delta(x)+0.9u(x)e^{-x}$?

So I need to integrate $\int_{-\infty}^{\infty}(.1\delta(x)+0.9u(x)e^{-x})xdx=\int_{0}^{\infty}(.1x\delta(x)+0.9xe^{-x})dx=\int_{0}^{\infty}.1x\delta(x)dx+\int_{0}^{\infty}0.9xe^{-x}dx=0.9$

Does it make sense? The average value is 0.9.... Strange...

Answer 1

Yes, that's fine. That is the integration for the expectation of this mixed distribution.