Computing conditional expectation of a function of $X$ given $X \leq t$

Question

I am interested in computing an expression of the form $\mathbb{E}[f(X)|X <= t]$ where $f$ is some continuous function and $X \sim \mathrm{Exp}(\lambda)$. I know I can write $$\mathbb{E}[f(X)] = \int_0^\infty f(s)\lambda e^{-\lambda s}ds$$ and using the law of total expectation I can get something like: $$\mathbb{E}[f(X)] = (1-e^{-\lambda t})\mathbb{E}[f(X)|X<=t] + e^{-\lambda t}\mathbb{E}[f(X)|X >t]$$ Is there a simple formula for the quantity I want in this setting?

Answer 1

In general, note that iterated expectations tells us

$$E[g(X)1_{X\leq t}]=E[E[g(X)1_{X\leq t}|1_{X\leq t}]]=E[g(X)|X\leq t]P(X\leq t),$$

and thus,

$$E[g(X)|X\leq t]={E[g(X)1_{X\leq t}] \over P(X\leq t)}\quad (1).$$

If $X$ has density $f$ and CDF $F$, then $(1)$ would be computed as

$$E[g(X)|X\leq t]={1 \over F(t)}{\int_{-\infty}^t g(x)f(x)dx}.$$

Answer 2

It looks like $E(f(X)|X\lt t)=\frac{\lambda\int\limits_0^t f(s)e^{-\lambda s}ds}{1-e^{-\lambda s}}$.