Expected Values of Continuous Random Variables

Question

Suppose I have a function $L(c)$ , where $c$ is a continuous random variable. Let

$L(c)=c+log(E[\theta])$

where $\theta$ is a continuous random variable and $E$ representing the corresponding expected value. My question is , is it fair to state that $L(c)$ itself is an expected value as it is expressed in terms of an $E(\cdot)$.

1. If this is true could you please explain why?

2. If this is not true could you please explain why?

Thank you in advance!

Answer 1

No. First of all, we generally denote random variables as capital letters $X$ and constants as lower case $x$ to avoid confusion.

$E\theta$ is just a constant. It's no longer random after taking expected value. So what this function $L(C)$ is doing is simply adding a constant to the random variable.

$L(C)$ is therefore still a random variable, except its now of the form

$C + a$

Where $a=\log E\theta$.

To refer to something as an "expected value" it needs to be of the form $E f(X)$ for some r.v. $X$ and function $f$. Since $L(C)$ is still a random variable (not constant) it is not an expected value.

Answer 2

It is not an expected value. Remember that a random variable is a function that maps sets to values. However, expectation is a function that maps functions to values. In your case. $L(c)$ still maps sets to values and the expectation has already resolved to a number.