Question

Let \(X\) be exponential with mean \(1 / \lambda ;\) that is, $$ f_{X}(x)=\lambda e^{-\lambda x}, \quad 01]\)

Short answer

The short answer is: \(E[X \mid X > 1] = \frac{e^{-\lambda + 1}}{\lambda - 1}\).

Step-by-step solution

Identify the conditional probability density function

To find the conditional probability density function (pdf), we use the following formula: $$ f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} $$ Since X is distributed exponentially and Y is the event (X > 1), we have $$ f_{X \mid X > 1}(x \mid x > 1) = \frac{f_X(x) \cdot I(x > 1)}{P(X > 1)} = \frac{\lambda e^{-\lambda x} \cdot I(x > 1)}{\int_{1}^{\infty} \lambda e^{-\lambda u} du} $$ Here, I(x > 1) is the indicator function that is equal to 1 when x > 1 and 0 otherwise.

Calculate the denominator of the conditional pdf

Now, we need to compute the lower bound of the integral in the denominator: $$ \int_{1}^{\infty} \lambda e^{-\lambda u} du = \left[-e^{-\lambda u}\right]_1^{\infty} = -e^{-\lambda \cdot \infty} + e^{-\lambda \cdot 1} = e^{-\lambda} $$

Compute the conditional pdf

Now, we can compute the conditional pdf: $$ f_{X \mid X > 1}(x \mid x > 1) = \frac{\lambda e^{-\lambda x} \cdot I(x > 1)}{e^{-\lambda}} = \lambda e^{-(\lambda - 1)x} \cdot I(x > 1) $$

Calculate the conditional expectation

Finally, we need to compute the conditional expectation \(E[X \mid X > 1]\): $$ E[X \mid X > 1] = \int_{\infty }^{\infty } x f_{X \mid X > 1}(x \mid x > 1) dx = \int_1^{\infty} x \lambda e^{-(\lambda - 1)x} dx $$ To solve this integral, we'll need to apply integration by parts: $$ u = x, \quad dv = \lambda e^{-(\lambda - 1)x} dx \\ du = dx, \quad v = - e^{-(\lambda - 1)x} $$ Now, using the integration by parts formula \(\int u dv = uv - \int v du\), we get: $$ E[X \mid X > 1] = -xe^{-(\lambda - 1)x}\bigg|_1^{\infty} + \int_1^{\infty} e^{-(\lambda - 1)x} dx = 0 + \frac{1}{\lambda - 1}\left[-e^{-(\lambda - 1)x}\right]_1^{\infty} = \frac{1}{\lambda - 1}(e^{-(\lambda - 1)} - e^{-\infty}) = \frac{e^{-\lambda + 1}}{\lambda - 1} $$ So, the expected value of X given that X > 1 is \(E[X \mid X > 1] = \frac{e^{-\lambda + 1}}{\lambda - 1}\).