The moment generating function (MGF) of a random variable X is a function that generates the moments of the distribution of X. The nth moment of X is given by the nth derivative of the MGF evaluated at t=0.

The mean (first moment) of X is given by the first derivative of the MGF evaluated at t=0. The variance of X is given by the second moment minus the square of the first moment.

First, let's find the first derivative of the MGF:

m'(t) = d/dt [(5+exp(2t))/6]
= (2/6) * exp(2t)

Evaluating this at t=0 gives the first moment (mean):

E(X) = m'(0) = (2/6) *

Question

The moment generating function (MGF) of a random variable X is a function that generates the moments of the distribution of X. The nth moment of X is given by the nth derivative of the MGF evaluated at t=0.

The mean (first moment) of X is given by the first derivative of the MGF evaluated at t=0. The variance of X is given by the second moment minus the square of the first moment.

First, let's find the first derivative of the MGF:

m'(t) = d/dt [(5+exp(2t))/6]
      = (2/6) * exp(2t)

Evaluating this at t=0 gives the first moment (mean):

E(X) = m'(0) = (2/6) *

Knowee AI · Accepted Answer

The moment generating function (MGF) of a random variable X is a function that generates the moments of the distribution of X. The nth moment of X is given by the nth derivative of the MGF evaluated at t=0.

The mean (first moment) of X is given by the first derivative of the MGF evaluated at t=0. The variance of X is given by the second moment minus the square of the first moment.

First, let's find the first derivative of the MGF:

m'(t) = d/dt [(5+exp(2t))/6]
      = (2/6) * exp(2t)

Evaluating this at t=0 gives the first moment (mean):

E(X) = m'(0) = (2/6) *

A random variable X has moment generating function m(t)=(5+exp(2t))/6. What is the variance of X?

Question

A random variable X has moment generating function $m(t) = \frac{5 + e^{2t}}{6}$ . What is the variance of X?

Solution

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