离散型几何分布f(x)=1/2^x的矩生成函数（MGF）求解咨询

Hey there! Let’s work through this problem together—you’ve already got the right starting point, so we just need to clear up the confusion around convergence and apply a key series formula.

Step 1: Rewrite the Series to Recognize Its Form #

First, let’s rearrange the terms in your MGF expression to make it easier to spot what kind of series we’re dealing with:
$$M(t)=\sum_{x=1}^\infty e^{tx} \frac{1}{2^x} = \sum_{x=1}^\infty \left( \frac{e^t}{2} \right)^x$$
This is a geometric series where each term is a multiple of the previous one, with common ratio $r = \frac{e^t}{2}$.

Step 2: Remember the Convergence Condition for Geometric Series #

Geometric series $\sum_{x=1}^\infty r^x$ converges only when $|r| < 1$. For our case, that means:
$$\left| \frac{e^t}{2} \right| < 1 \implies e^t < 2 \implies t < \ln 2$$
So your initial thought about divergence isn’t wrong for all values of $t$—the MGF only exists (i.e., the series converges) when $t < \ln 2$. Outside this range, the series does diverge, but within it, we can sum it up.

Step 3: Apply the Geometric Series Sum Formula #

For a convergent geometric series starting at $x=1$, the sum is:
$$\sum_{x=1}^\infty r^x = \frac{r}{1 - r}$$
Plugging in our $r = \frac{e^t}{2}$:
$$M(t) = \frac{\frac{e^t}{2}}{1 - \frac{e^t}{2}}$$
Now simplify the denominator by getting a common denominator:
$$1 - \frac{e^t}{2} = \frac{2 - e^t}{2}$$
Substitute back in, and the 2 in the numerator and denominator cancels out:
$$M(t) = \frac{e^t}{2 - e^t}$$
Which matches the expected result!

Quick Recap #

• The series is a geometric series with ratio $r = e^t/2$
• It converges only when $t < \ln 2$ (this is the domain where the MGF is defined)
• Using the geometric series sum formula gives you the simplified expression you’re looking for

内容的提问来源于stack exchange，提问作者Grak