Conditional expectation of X given X > 1

Question

Let X ~ Exp(λ). How would we find E(X|X > 1).

I know that the E(X|X > 1) = $\int x P(X | X > 1)$ = $\int x \frac{P(X = x \space and \space X>1)}{P(X > 1)}$

But im not sure how to evaluate the fraction.

Thanks!

Answer 1

• Exponential distribution is continuous.

\begin{align} E[X|X>1] &= \int_1^\infty x \frac{\lambda \exp(-\lambda x)}{\exp(-\lambda )} \, dx \\ &=\int_1^\infty x \lambda \exp(-\lambda (x-1)) \, dx\\ &= \int_0^\infty (y+1) \lambda\exp(-\lambda y) \, dy\\ &= \int_0^\infty \lambda\exp(-\lambda y) \, dy + \int_0^\infty y \lambda \exp(-\lambda y) \, dy \\ &= 1+E[X]\end{align}

This is known as memoryless property of exponential distribution.

Answer 2

Perhaps a simpler approach is to note that $$E(X\mid X\gt1)=1+E(X)$$

since the exponential distribution is memoryless.

As Vincent pointed out, the exponential distribution is continuous so you should be integrating.

We have $$E(X)=\int_0^{\infty}x\lambda e^{-\lambda x}=\frac{1}{\lambda}$$

Answer 3

What you're trying for is: $~\mathsf E(X\mid X>1) ~{= \dfrac{\mathsf E(X\,\mathbf 1_{X>1})}{\mathsf E(\mathbf 1_{X>1})} \\= \dfrac{\int_1^\infty \lambda x e^{-\lambda d x}\mathsf d x}{\int_1^\infty \lambda e^{-\lambda d x}\mathsf d x}\\~~\vdots\\=1+\lambda^{-1}}$

Answer 4

You have that the expectation is $$\mathbb{E}[X|X>1]=\int_0^{\infty}\mathbb{P}(X>a|X>1)da=\int_{0}^{\infty}\frac{\mathbb{P}(X>a, X>1)}{\mathbb{P}( X>1)}da=\frac{1}{e^{-\lambda}}(\int_{0}^{1}\mathbb{P}(X>1)da+\int_{1}^{\infty}\mathbb{P}(X>a)da)=\frac{1}{e^{-\lambda}}([ae ^{-\lambda}]^1_0+[\frac{e^{-\lambda a}}{\lambda}]^{\infty}_1)=\frac{1}{e^{-\lambda}}(e^{-\lambda}+\frac{e^{-\lambda a}}{\lambda})=1+\frac{1}{\lambda}$$