MFG - Probability .?

Question

Let Y~exp(1) , X~min(Y,1) .
Compute MFG of X.

Answer 1

I'm not 100% sure what you're asking, but I think it's this:
Let X be a random variable defined by X = min(Y, 1) where Y ~ exp(1). Compute the MGF of X (not MFG, which I spent some time trying to track down!)

We know y has a pdf (probability density function) of f(y) = e^(-x) for y >= 0, 0 for y < 0. Also, the cmf for y is
F(y) = 0, y < 0; 1 - e^(-y), y >= 0.
So x will have the same pdf and cmf for x < 1, but then at x = 1 we will get a lump sum probability corresponding to the probability that Y >= 1. This is 1 - F(1) = e^(-1) = 1/e. For x > 1 the pdf is 0.

The MGF of X is given by M(t) = E(e^(tX)) for t ∈ R. We can split this out into the lump-sum part and the integral part:
M(t) = P(X=1) e^t + ∫(-∞ to ∞) f(x) e^(tx) dx
= 1/e e^t + ∫(0 to 1) e^(-x) e^(tx) dx
= e^(t-1) + ∫(0 to 1) e^((t-1)x) dx
= e^(t-1) + [e^((t-1)x) / (t-1)] [0 to 1]
= e^(t-1) + [e^(t-1) / (t-1) - e^(0) / (t-1)]
= e^(t-1) (1 + 1/(t-1)) - 1/(t-1)
= e^(t-1) (t / (t-1)) - 1/(t-1)
= (t e^(t-1) - 1) / (t-1).

Answer 2

I just reviewed the definition of the mgf. It's defined in terms of the expectation of a function of the RV. I'd recommend that you review the definition, and note that integrals can be broken up into pieces during evaluation. Since and exponential is part of the definition, I think that you should make some progress....

Answer 3

Love to help, but don't understand your notation and abbreviations. What's the ~ about? What's MFG?