If Y has a binomial distribution with n trials and probability of success p, show that the
moment-generating function for Y is
m(t) = (pe^t + q)^n , where q = 1 − p

To derive the moment-generating function (MGF) for a binomially distributed random variable \( Y \), we need to follow these steps.

### 1. Definition of the Moment-Generating Function (MGF)
The moment-generating function \( M_Y(t) \) of a random variable \( Y \) is defined as:

\[
M_Y(t) = \mathbb{E}[e^{tY}]
\]

where \( \mathbb{E} \) denotes the expected value.

### 2. Binomial Distribution Setup
The random variable \( Y \) follows a binomial distribution with parameters \( n \) (the number of trials) and \( p \) (the probability of success in each trial). The probability mass function (PMF) for \( Y \) is:

\[
P(Y = k) = \binom{n}{k} p^k (1 - p)^{n-k}, \quad k = 0, 1, 2, \dots, n
\]

### 3. Moment-Generating Function for the Binomial Distribution
Using the definition of the MGF, we write:

\[
M_Y(t) = \mathbb{E}[e^{tY}] = \sum_{k=0}^{n} e^{tk} P(Y = k)
\]

Substitute the PMF of the binomial distribution into this sum:

\[
M_Y(t) = \sum_{k=0}^{n} e^{tk} \binom{n}{k} p^k (1 - p)^{n-k}
\]

Now, factor out terms that don't depend on \( k \):

\[
M_Y(t) = \sum_{k=0}^{n} \binom{n}{k} \left(p e^t \right)^k (1 - p)^{n-k}
\]

### 4. Recognizing the Binomial Theorem
The expression inside the sum resembles the binomial expansion:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}
\]

Set \( a = p e^t \) and \( b = 1 - p \). Thus, the sum becomes:

\[
M_Y(t) = \left( p e^t + (1 - p) \right)^n
\]

### 5. Simplification
Since \( q = 1 - p \), the final expression for the moment-generating function is:

\[
M_Y(t) = \left( p e^t + q \right)^n
\]

This completes the derivation of the moment-generating function for a binomially distributed random variable \( Y \).

### Summary
The moment-generating function for \( Y \) is:

\[
M_Y(t) = \left( p e^t + q \right)^n
\]

where \( q = 1 - p \).

---

Would you like more details on any part of this explanation, or do you have any questions?

Here are some related questions to explore further:

1. How does the MGF relate to the mean and variance of a binomial distribution?
2. Can the MGF be used to find higher moments (e.g., skewness or kurtosis)?
3. What is the MGF for other discrete distributions, such as the Poisson distribution?
4. How would you derive the cumulative distribution function (CDF) from the binomial PMF?
5. How does the moment-generating function help in proving the Central Limit Theorem for binomial distributions?

**Tip:** The moment-generating function is useful because it helps in calculating moments and analyzing the distribution’s behavior via its derivatives at \( t = 0 \).

If Y has a binomial distribution with n trials and probability of success p, show that the moment-generating function for Y is m(t) = (pe^t + q)^n, where q = 1 − p.

Learn how to derive the moment-generating function (MGF) for a binomially distributed random variable Y with n trials and probability of success p. This detailed solution walks through the key formulas and concepts, including the binomial distribution's PMF, the binomial theorem, and the properties of MGFs.

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