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If a random variable X assumes the value of a Bernoulli(1/3) random variable with a probability of 1/4, and assumes the value of a Uniform(0,1) random variable with a probability of 3/4, how does one obtain a moment generating function of X?

I'm fuzzy on how exactly mixed distributions work, and I'm not sure that term even applies here, however I came across this problem and thought it to be like no other.

In any case, I believe the moment generating function for the uniform part of this variable would be ((e^t)-1)/t. And the Bernoulli mgf would be 2/3+(1/3)(e^t).

Given that the Bernoulli variable occurs with probability 3/4 and the Uniform variable with probability 1/4, how does one combine these into one cohesive mgf? Have I simply forgotten a formula?

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The moment generating function is an expectation value. Particularly, it's $E(e^{tX}),$ so the question is how to take this expectation value when $X$ is a mixture of two distributions.

But you can think of a mixture as an initial coinflip which determines which conditional distribution the variable comes from. So by total probability we have $$E(e^{tX}) = E(e^{tX}|Bernoulli)P(Bernoulli) + E(e^{tX}|Uniform)P(Uniform).$$

In other words, it's the weighted sum of the two MGFs.

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  • $\begingroup$ Thank you, I suspected it might be something this simple, but wasn't sure. $\endgroup$ – Bad at algebra and proofs Jun 16 '17 at 23:10
  • $\begingroup$ Yeah, but you're right to not assume... a lot of times it isn't. $\endgroup$ – spaceisdarkgreen Jun 16 '17 at 23:19

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