If $X$ is an IID exponential random variable with parameter $\lambda = 1/3$, what is $E[X|X \ge 3]$?

Question

If $X$ is an IID exponential random variable with parameter $\lambda = 1/3$, what is $E[X|X \ge 3]$?

How is $E[X|X \ge 3]$ defined? $E[X] = \int_{0}^{\infty} f_X(x)dx$, but what does $E[X|X \ge a]$ mean?

Answer 1

Analogously to how $\mathbb P(A \mid B)$ is defined as $\frac{\mathbb P(A \cap B)}{\mathbb P(B)}$, the conditional expectation $\mathbb E[X \mid A]$ is defined as $ \frac{\mathbb E[X \cdot \mathbf 1_A]}{\mathbb P(A)}$, where $\mathbf 1_A$ denotes the indicator function of event $A$ (i.e. the random variable that is $1$ when event $A$ occurs and is $0$ when $A$ does not occur). In this case, the numerator of that expression would be computed as $$\mathbb E[X \cdot \mathbf 1_{\{X \geq 3\}}] = \int_0^{\infty} x \cdot \mathbf 1_{\{X \geq 3\}} \cdot f(x) \, \textrm d x = \int_3^{\infty} x \cdot f(x)\, \textrm d x$$ where $f(x)$ is the usual exponential density function.