# How to calculate conditional expectation E[g(x) | x>= Q] for x ~ exp(1)

• MHB
• user_01

#### user_01

Given that $X$ is exponentially distributed continuous random variable $X\sim \exp(1)$ and $g(x)$ is as below. How can I find the Expectectaion of $g(x)$ for the condition that $x\geq Q$, i.e. $\mathbb{E}[g(x)\ | \ x\geq Q]$.

$$g(x) = \frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big)$$

I suppose I should start by considering an event $\Phi$ such that $\Phi = \mathbb{P}[x \geq Q]$. However, I don't know how to go around this condition.

All constants are positive real values.

Is the following solution correct for the above question? If it is OK, then I have found the solution. But I will really appreciate if someone can let me know if the following method is correct.

$$\mathbb{E}[g[x]] = \int_Q^\infty g(x). f(x) dx$$

$$\mathbb{E}[g[x]]= \int_Q^\infty\frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big) e^{-x} dx$$

Hi user_01,

Yes, your proposed solution is correct. The relevant theorem here is known as the Law of the Unconscious Statistician (LOTUS). See here.