The lifetime (in thousands of hours) X of a light bulb has pdf g(x)= \lambda e^{-\lambda x}, \hspace{3mm} x\geq 0 for some unknown \lambda >0. Find \mathbb E[X].

Question

\mathbb E[X]=\quad

Answer 1

To find the expected value of X, we can use the formula:

E[X] = ∫ x * g(x) dx

Plugging in the given pdf g(x) = λe^(-λx), we have:

E[X] = ∫ x * λe^(-λx) dx

To solve this integral, we can use integration by parts. Let u = x and dv = λe^(-λx) dx. Then, du = dx and v = -e^(-λx)/λ.

Using the formula for integration by parts:

∫ u * dv = uv - ∫ v * du

We have:

E[X] = [-x * e^(-λx)/λ] - ∫ (-e^(-λx)/λ) dx
= [-x * e^(-λx)/λ] + [e^(-λx)/λ^2] + C

where C is the constant of integration.

To find the limits of integration, recall that the pdf g(x) is defined for x ≥ 0. Therefore, the integral is evaluated from 0 to ∞.

E[X] = [0 * e^(-λ*0)/λ] + [e^(-λ*0)/λ^2] - [-∞ * e^(-λ*∞)/λ] - [e^(-λ*∞)/λ^2]

Since e^(-λ∞) approaches 0 as λ approaches infinity, the last two terms in the equation above simplify to 0.

E[X] = 0 + (1/λ^2) - 0 - 0
= 1/λ^2

Therefore, the expected value of X is 1/λ^2.