Question 1169911: A fair coin is tossed n times. Let X be the number of times that a head
occurs in the n tosses.
Determine the moment generating function of X.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's determine the moment generating function (MGF) of X, where X is the number of heads in n tosses of a fair coin.
**1. Understand the Distribution of X:**
Since the coin is fair, the probability of getting a head in a single toss is p = 1/2. Each toss is independent. Therefore, X follows a binomial distribution with parameters n (number of trials) and p = 1/2 (probability of success).
X ~ Binomial(n, 1/2)
**2. Recall the PMF of a Binomial Distribution:**
The probability mass function (PMF) of a binomial distribution is given by:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
In our case, p = 1/2, so:
P(X = k) = (n choose k) * (1/2)^k * (1/2)^(n - k)
P(X = k) = (n choose k) * (1/2)^n
**3. Define the Moment Generating Function (MGF):**
The MGF of a discrete random variable X is defined as:
M_X(t) = E[e^(tX)] = Σ e^(tk) * P(X = k)
where the summation is taken over all possible values of k.
**4. Apply the Definition to Our Binomial Distribution:**
M_X(t) = Σ[k=0 to n] e^(tk) * (n choose k) * (1/2)^n
M_X(t) = (1/2)^n * Σ[k=0 to n] (n choose k) * e^(tk)
M_X(t) = (1/2)^n * Σ[k=0 to n] (n choose k) * (e^t)^k * 1^(n - k)
**5. Recognize the Binomial Theorem:**
The binomial theorem states:
(a + b)^n = Σ[k=0 to n] (n choose k) * a^k * b^(n - k)
In our case, a = e^t and b = 1. Therefore:
Σ[k=0 to n] (n choose k) * (e^t)^k * 1^(n - k) = (e^t + 1)^n
**6. Substitute Back into the MGF:**
M_X(t) = (1/2)^n * (e^t + 1)^n
M_X(t) = [(e^t + 1) / 2]^n
**Therefore, the moment generating function of X is:**
M_X(t) = [(e^t + 1) / 2]^n
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