Conditional expectation of exponential RV with condition on itself

probability

The lifetime of a machine part is exponentially distributed with a mean of five years. Calculate the mean >lifetime of the part, given that it survives less than ten years.

Correct Answer: 3.425

My work:

X ~ exp($\lambda = 1/5$), Want to calcuate

$E[X|X\leq 10] = \int_0^\infty xP(X|X\leq 10)dx = \int_0^\infty x\frac{P(X=x, X\leq 10)}{P(X\leq 10)}dx = \frac{\int_0^{10} xP(X=x)dx}{P(X\leq 10)}?$ Is that last step valid? I think it makes sense intuitively but not sure if the math/reasoning behind is right..

Best Answer

By definition

$$E(X|A)=\frac{E(X 1_A)}{P(A)}$$ Conditional_expectation_with_respect_to_an_event

choose $A=\{ X\leq 10\}$ and

$$E(X|X\leq 10)=\frac{E(X 1_{X\leq 10})}{P(X\leq 10)}= \frac{\int x 1_{x\leq 10} f(x) dx}{P(X\leq 10)} = \frac{\int_0^{10} x f(x) dx}{P(X\leq 10)}$$

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