技术求助：如何从给定概率质量函数（PMF）推导矩生成函数（MGF）

Let's break this down step by step—no skipped steps, promise. I know you already have the first piece (multiplying $e^{tx}$ by the PMF), so let's pick up right there and work through the rest.

Step 1: Recall the MGF Definition #

First, remember that for a discrete random variable $X$, the moment generating function $M_X(t)$ is defined as:
$$M_X(t) = E\left[e^{tX}\right] = \sum_{\text{all } x} e^{tx} f(x)$$
Your PMF applies when $X \geq 2$, so $x$ takes values 2, 3, 4, ... (all integers starting at 2).

Step 2: Write Out the Summation #

Substitute your PMF $f(x) = \frac{1}{2^{x-1}}$ into the MGF formula:
$$M_X(t) = \sum_{x=2}^{\infty} e^{tx} \cdot \frac{1}{2^{x-1}}$$

Step 3: Rewrite the PMF to Simplify the Sum #

Let's rearrange the term $\frac{1}{2^{x-1}}$ to make it easier to spot a geometric series. Notice that:
$$\frac{1}{2^{x-1}} = \frac{2}{2^x} = 2 \cdot \left(\frac{1}{2}\right)^x$$
Substitute this back into the summation:
$$M_X(t) = 2 \sum_{x=2}^{\infty} e^{tx} \cdot \left(\frac{1}{2}\right)^x = 2 \sum_{x=2}^{\infty} \left( \frac{e^t}{2} \right)^x$$

Step 4: Use the Geometric Series Formula #

For a geometric series $\sum_{x=k}^{\infty} r^x$, if $|r| < 1$, the sum equals $\frac{r^k}{1 - r}$. Here:

• Our $r = \frac{e^t}{2}$
• Our starting index $k = 2$

First, we need the convergence condition: $|r| < 1$ translates to $\frac{e^t}{2} < 1$ (since $e^t$ is always positive), so $e^t < 2$ or $t < \ln 2$. The MGF only exists for $t$ in this interval.

Now apply the formula:
$$M_X(t) = 2 \cdot \frac{\left( \frac{e^t}{2} \right)^2}{1 - \frac{e^t}{2}}$$

Step 5: Simplify the Expression #

Let's simplify the numerator and denominator separately:

1. Numerator: $2 \cdot \left( \frac{e^{2t}}{4} \right) = \frac{e^{2t}}{2}$
2. Denominator: $1 - \frac{e^t}{2} = \frac{2 - e^t}{2}$

Divide the numerator by the denominator (the 2s cancel out):
$$M_X(t) = \frac{e^{2t}}{2 - e^t}, \quad t < \ln 2$$

Quick Check #

To make sure we didn't make a mistake, plug in $t=0$ (the MGF should always equal 1 at $t=0$):
$$M_X(0) = \frac{e^{0}}{2 - e^0} = \frac{1}{2-1} = 1$$
Perfect, that checks out.

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