Show that when X follows an exponential distribution with parameter ,E.X n / D nŠ= n, n  0

It seems like there's a bit of confusion in your question as it contains some special characters that are not clearly defined. However, I'll provide a general solution for the expected value of an exponential distribution.

The exponential distribution is defined as:

f(x|λ) = λe^(-λx) for x >= 0, and 0 otherwise.

Where:

• λ > 0 is the rate parameter
• x is a random variable

The expected value (E[X]) of an exponential distribution is calculated as:

E[X] = ∫_0^∞ xλe^(-λx) dx

To solve this integral, we can use integration by parts, where:

u = x, dv = λe^(-λx) dx du = dx, v = -e^(-λx)

According to the integration by parts formula ∫udv = uv - ∫vdu, we get:

E[X] = [-xe^(-λx)]_0^∞ + ∫_0^∞ e^(-λx) dx

The first term of the right hand side is zero (you can confirm this by substituting the limits of x). The second term is the integral of an exponential distribution, which is equal to 1/λ.

So, the expected value E[X] = 1/λ.

This shows that the expected value of an exponential distribution with rate parameter λ is 1/λ. If you meant something different with your question, please provide additional details or clarify the special characters.