Find the variance of the moment generating function Mx(t)=1+2t+4t^2+8t^3...

The moment generating function (MGF) you provided appears to be a geometric series with a common ratio of 2t. The sum of a geometric series is given by a / (1 - r), where a is the first term and r is the common ratio. In this case, a = 1 and r = 2t, so the MGF is M(t) = 1 / (1 - 2t).

The variance of a random variable X can be found using the formula Var(X) = E[X^2] - (E[X])^2, where E[X] is the expected value of X and E[X^2] is the expected value of X squared. These can be found using the MGF as follows:

E[X] = M'(0) = d/dt M(t) at t = 0 E[X^2] = M''(0) = d^2/dt^2 M(t) at t = 0

First, we find the first derivative of M(t):

M'(t) = d/dt (1 / (1 - 2t)) = 2 / (1 - 2t)^2

Evaluating at t = 0 gives E[X] = M'(0) = 2.

Next, we find the second derivative of M(t):

M''(t) = d^2/dt^2 (1 / (1 - 2t)) = 8 / (1 - 2t)^3

Evaluating at t = 0 gives E[X^2] = M''(0) = 8.

Finally, we substitute these into the formula for the variance:

Var(X) = E[X^2] - (E[X])^2 = 8 - 2^2 = 4.