Let's revisit the calculation of the mean ($\mu$) using the moment-generating function (MGF).

MGF $M_X(t)$ Revisited

The MGF is given by:

$M_X(t) = \frac{e^t}{3 - 2e^t}$

1. Calculate the Mean ($\mu$)

The mean of $X$, $\mu = \mathbb{E}[X]$, can be found by taking the first derivative of the MGF with respect to $t$ and evaluating it at $t = 0$:

$\mu = M'_X(0)$

First, we'll find the derivative $M'_X(t)$.

2. Derivative of the MGF

The derivative of $M_X(t)$ with respect to $t$ is:

$M'_X(t) = \frac{d}{dt} \left(\frac{e^t}{3 - 2e^t}\right)$

We'll use the quotient rule, which states that if $f(t) = \frac{g(t)}{h(t)}$, then:

$f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2}$

Here, $g(t) = e^t$ and $h(t) = 3 - 2e^t$. Their derivatives are:

$g'(t) = e^t$ $h'(t) = -2e^t$

Now, applying the quotient rule:

$M'_X(t) = \frac{e^t(3 - 2e^t) - e^t(-2e^t)}{(3 - 2e^t)^2}$

Simplifying:

$M'_X(t) = \frac{3e^t - 2e^{2t} + 2e^{2t}}{(3 - 2e^t)^2} = \frac{3e^t}{(3 - 2e^t)^2}$

3. Evaluate at $t = 0$ to Find the Mean

Now, we substitute $t = 0$ into the derivative:

$\mu = M'_X(0) = \frac{3e^0}{(3 - 2e^0)^2} = \frac{3 \times 1}{(3 - 2 \times 1)^2} = \frac{3}{1^2} = 3$

Correct Mean

The correct mean $\mu$ is $\mu = 3$.

Summary:

• Mean $\mu = 3$.

Let me know if you have any further questions or need additional details!

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Here are five related questions:

1. How would the calculation of the mean change for a different PMF?
2. What is the physical or probabilistic interpretation of the mean?
3. Can the MGF be used to calculate other moments like the skewness?
4. How would you calculate the mean directly from the PMF without using the MGF?
5. What is the relationship between the MGF and the characteristic function of a random variable?

Tip: Double-checking derivative calculations is crucial in problems involving MGFs to ensure accurate results for the mean and variance.