Getting the probability of X from a moment generating function

Question

I'm trying to understand moment generating functions, and I understand how to find the mean, variance, and standard deviation, as well as the distribution from the m.g.f., but I don't know how to find the probability function from the m.g.f.

I have a function given:

M(t) = (0.3 + 0.7e^t)⁵

I know that the distribution is uniform, and is given as

b(5, 0.7)

so n=5.

I was able to find the mean and variance by using derivatives:

M'(t) = 0.84035e^t * (0.428571 + e^t)⁴

M''(t) = 4.20175e^t * (0.0857142 + e^t) * (0.428571 + e^t)³

so μ = 3.5 and σ² = 1.05

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I now need to find the probability that 1 ≤ X ≤ 2. I know the formula for the m.g.f.

M(t) = ∑ e^tx * p(x) for all x.

and I know the domain is -2 ≤ X ≤ 2.

I don't know where to go from here.

Answer 1

In general, there are no trivial ways to recover the CDF or PDF of a distribution from its MGF. The MGF is related to the Laplace transform of the PDF, and generally requires contour integration for recovery.

For your specific example, by pattern matching it is clear that $0.3 + 0.7e^t$ is the MGF of a Bernoulli random variable with probability $p = 0.7$ of success. Then consider the sum $X_1 + \cdots + X_5$ where $X_1, \dots, X_5 \stackrel{iid}{\sim} \text{Bern}(0.7)$. Try computing the MGF of this sum, and you'll find that it is exactly the MGF you are given.

In general, adding random variables that are iid results in an MGF taken to a power of the individual MGF.