Math Problem Statement

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

Solution

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

1. Definition of the Moment-Generating Function (MGF)

The moment-generating function MY(t)M_Y(t) of a random variable YY is defined as:

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

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

2. Binomial Distribution Setup

The random variable YY follows a binomial distribution with parameters nn (the number of trials) and pp (the probability of success in each trial). The probability mass function (PMF) for YY is:

P(Y=k)=(nk)pk(1p)nk,k=0,1,2,,nP(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:

MY(t)=E[etY]=k=0netkP(Y=k)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:

MY(t)=k=0netk(nk)pk(1p)nkM_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 kk:

MY(t)=k=0n(nk)(pet)k(1p)nkM_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=k=0n(nk)akbnk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}

Set a=peta = p e^t and b=1pb = 1 - p. Thus, the sum becomes:

MY(t)=(pet+(1p))nM_Y(t) = \left( p e^t + (1 - p) \right)^n

5. Simplification

Since q=1pq = 1 - p, the final expression for the moment-generating function is:

MY(t)=(pet+q)nM_Y(t) = \left( p e^t + q \right)^n

This completes the derivation of the moment-generating function for a binomially distributed random variable YY.

Summary

The moment-generating function for YY is:

MY(t)=(pet+q)nM_Y(t) = \left( p e^t + q \right)^n

where q=1pq = 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=0t = 0.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Moment-Generating Function
Expected Value

Formulas

Moment-Generating Function (MGF): M_Y(t) = E[e^{tY}]
Binomial PMF: P(Y = k) = (n choose k) * p^k * (1-p)^(n-k)
Binomial Theorem: (a + b)^n = sum_{k=0}^{n} (n choose k) * a^k * b^{n-k}

Theorems

Binomial Theorem
Properties of Moment-Generating Functions

Suitable Grade Level

College/University Level