\begin{tabular}{l|l} A2. Let $$ X \sim \operatorname{Po}(\mu) $$. \ (a) Find the moment generating function $$ M_{X}(t) $$. & {$$ [6] $$} \ (b) Verify that $$ E(X)=\operatorname{Var}(X)=\mu $$ & {$$ [6] $$}\end{tabular}

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**(a) Finding the moment generating function $$ M_X(t) $$:**

The moment generating function (MGF) of a random variable $$ X $$ is defined as  
$$
M_X(t) = E\left(e^{tX}\right)
$$
For $$ X \sim \operatorname{Po}(\mu) $$, the probability mass function is  
$$
P(X=k) = \frac{e^{-\mu}\mu^k}{k!}, \quad k=0,1,2,\ldots
$$
Thus, we have
$$
M_X(t)= \sum_{k=0}^\infty e^{tk} \frac{e^{-\mu}\mu^k}{k!} = e^{-\mu}\sum_{k=0}^\infty \frac{(\mu e^t)^k}{k!}
$$
Recognize that the series  
$$
\sum_{k=0}^\infty \frac{(\mu e^t)^k}{k!} = e^{\mu e^t}
$$
So,
$$
M_X(t) = e^{-\mu} e^{\mu e^t} = e^{\mu(e^t-1)}
$$

**(b) Verifying that $$ E(X)=\operatorname{Var}(X)=\mu $$:**

1. **Finding $$ E(X) $$:**

The first moment can be obtained by differentiating the MGF:
   $$
   E(X) = M'_X(0)
   $$
   Start with
   $$
   M_X(t) = e^{\mu(e^t-1)}
   $$
   Differentiate with respect to $$ t $$:
   $$
   M'_X(t) = e^{\mu(e^t-1)} \cdot \mu e^t
   $$
   Evaluating at $$ t=0 $$:
   $$
   M'_X(0) = e^{\mu(e^0-1)} \cdot \mu e^0 = e^{\mu(1-1)} \cdot \mu = \mu
   $$
   Therefore,
   $$
   E(X)=\mu.
   $$

2. **Finding $$ \operatorname{Var}(X) $$:**

The second moment is given by
   $$
   E(X^2)= M''_X(0)
   $$
   Differentiate $$ M'_X(t) $$ with respect to $$ t $$ to get the second derivative:
   $$
   M'_X(t)= e^{\mu(e^t-1)} \cdot \mu e^t
   $$
   Differentiate using the product rule:
   $$
   M''_X(t)= \frac{d}{dt}\left(e^{\mu(e^t-1)}\right) \cdot \mu e^t + e^{\mu(e^t-1)} \cdot \mu e^t
   $$
   The derivative of $$ e^{\mu(e^t-1)} $$ is:
   $$
   \frac{d}{dt}\left(e^{\mu(e^t-1)}\right) = e^{\mu(e^t-1)} \cdot \mu e^t
   $$
   Therefore,
   $$
   M''_X(t)= e^{\mu(e^t-1)} \cdot \mu e^t \cdot \mu e^t + e^{\mu(e^t-1)} \cdot \mu e^t = e^{\mu(e^t-1)}\mu e^t\left(\mu e^t+1\right)
   $$
   Evaluate at $$ t=0 $$:
   $$
   M''_X(0)= e^{\mu(e^0-1)}\mu e^0\left(\mu e^0+1\right) = e^{\mu(1-1)}\mu\left(\mu+1\right)=\mu\left(\mu+1\right)
   $$
   Now, the variance is given by
   $$
   \operatorname{Var}(X)= E(X^2)-\left(E(X)\right)^2 = \mu(\mu+1) - \mu^2 = \mu.
   $$

Thus, we have verified that:
$$
E(X)=\mu \quad \text{and} \quad \operatorname{Var}(X)=\mu.
$$
The moment generating function $$ M_X(t) $$ for $$ X \sim \operatorname{Po}(\mu) $$ is $$ e^{\mu(e^t - 1)} $$. The expected value $$ E(X) $$ and variance $$ \operatorname{Var}(X) $$ are both equal to $$ \mu $$.

A2. Let .

(a) Find the moment generating function .

(b) Verify that

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