已知N~泊松(λ=6)、X|N=n~二项(n,0.6)，求(N,X)联合矩生成函数遇阻

Hey there! Let's break down how to find the joint moment generating function for (N,X) without getting tangled up in that messy double sum. Here's a step-by-step walkthrough:

First, remember that the joint moment generating function (MGF) is defined as:
$$ M_{N,X}(s,t) = \mathbb{E}\left[e^{sN + tX}\right] $$
Instead of jumping straight into summing over the joint PMF, we can use the law of total expectation to simplify things—this is the key to avoiding that tricky double sum.

We can rewrite the joint MGF by conditioning on N first:
$$ M_{N,X}(s,t) = \mathbb{E}\left[ \mathbb{E}\left[e^{sN + tX} \mid N\right] \right] $$
When we condition on ( N = n ), ( e^{sN} ) becomes ( e^{sn} ) (a constant with respect to X), so we can factor that out of the inner expectation:
$$ \mathbb{E}\left[e^{sN + tX} \mid N=n\right] = e^{sn} \cdot \mathbb{E}\left[e^{tX} \mid N=n\right] $$

We know that given ( N=n ), X follows a Binomial(n, 0.6) distribution. The MGF of a Binomial(n,p) random variable is ( (1-p + pe^t)n ). Plugging in p=0.6, this gives:
$$ \mathbb{E}\left[e^{tX} \mid N=n\right] = (0.4 + 0.6e^t)n $$
Substitute this back into our expression:
$$ M_{N,X}(s,t) = \mathbb{E}\left[ e^{sn} \cdot (0.4 + 0.6e^t)n \right] = \mathbb{E}\left[ \left( e^s(0.4 + 0.6e^t) \right)^N \right] $$

N follows a Poisson distribution with λ=6, whose MGF is ( \mathbb{E}\left[e^{uN}\right] = e^{\lambda(eu - 1)} ). Notice that our current expression is exactly the Poisson MGF where ( u = \ln\left(e^s(0.4 + 0.6e^t)\right) = s + \ln(0.4 + 0.6e^t) ).

Substitute u into the Poisson MGF and simplify:
$$ M_{N,X}(s,t) = e^{6\left( e^{s + \ln(0.4 + 0.6e^t)} - 1 \right)} = e^{6\left( e^s(0.4 + 0.6e^t) - 1 \right)} $$

That's it! By using total expectation instead of directly summing the joint PMF, we avoid the messy double sum and get a clean closed-form solution.

内容的提问来源于stack exchange，提问作者EllipticalInitial