Understanding Moment Generating Functions for Discrete Variables

The Moment Generating Function (MGF), denoted by $M_X(t)$, is a powerful tool in probability theory used to find the moments (like mean, variance, etc.) of a random variable. For a discrete random variable X, the MGF is defined as $M_X(t) = E(e^{tX})$.

The given moment generating function is $M_X(t) = (q + pe^t)^n$. This specific form of MGF is characteristic of a discrete random variable following a Binomial distribution with parameters n (number of trials) and p (probability of success). Here, q represents the probability of failure, so $q = 1 - p$.

Calculating Expected Value $E(X^2)$ using MGF

The moments of a random variable can be found by differentiating the MGF and evaluating it at $t=0$. Specifically:

• $E(X) = M_X'(0)$ (The first moment, or mean)
• $E(X^2) = M_X''(0)$ (The second moment about the origin)
• And generally, $E(X^k) = M_X^{(k)}(0)$

Our goal is to find $E(X^2)$, which requires us to compute the second derivative of $M_X(t)$ with respect to $t$ and then substitute $t=0$.

Step 1: Find the First Derivative of the MGF

Given $M_X(t) = (q + pe^t)^n$. We differentiate $M_X(t)$ with respect to $t$ using the chain rule:

$\frac{d}{dt} (u^n) = nu^{n-1} \frac{du}{dt}$

Here, $u = q + pe^t$. So, $\frac{du}{dt} = \frac{d}{dt}(q + pe^t) = 0 + p \frac{d}{dt}(e^t) = pe^t$.

Therefore, the first derivative $M_X'(t)$ is:

$M_X'(t) = n(q + pe^t)^{n-1} \cdot (pe^t)$

$M_X'(t) = npe^t (q + pe^t)^{n-1}$

Step 2: Find the Second Derivative of the MGF

Now, we differentiate $M_X'(t) = npe^t (q + pe^t)^{n-1}$ with respect to $t$ using the product rule: $(uv)' = u'v + uv'$.

Let $u = npe^t$ and $v = (q + pe^t)^{n-1}$.

First, find the derivatives of u and v with respect to t:

• $u' = \frac{d}{dt}(npe^t) = npe^t$
• $v' = \frac{d}{dt}((q + pe^t)^{n-1})$

Using the chain rule for $v'$ (similar to Step 1, but with power $n-1$):

$v' = (n-1)(q + pe^t)^{(n-1)-1} \cdot \frac{d}{dt}(q + pe^t)$

$v' = (n-1)(q + pe^t)^{n-2} \cdot (pe^t)$

$v' = p(n-1)e^t (q + pe^t)^{n-2}$

Now, apply the product rule $M_X''(t) = u'v + uv'$:

$M_X''(t) = (npe^t)(q + pe^t)^{n-1} + (npe^t) \cdot [p(n-1)e^t (q + pe^t)^{n-2}]$

$M_X''(t) = npe^t (q + pe^t)^{n-1} + np^2(n-1)e^{2t} (q + pe^t)^{n-2}$

Step 3: Evaluate the Second Derivative at $t=0$

To find $E(X^2)$, we substitute $t=0$ into the expression for $M_X''(t)$:

$E(X^2) = M_X''(0) = npe^0 (q + pe^0)^{n-1} + np^2(n-1)e^{2 \cdot 0} (q + pe^0)^{n-2}$

Recall that $e^0 = 1$ and $e^{2 \cdot 0} = e^0 = 1$. Also, for a Binomial distribution, $q+p=1$. Substitute these values:

$E(X^2) = np(1)(q + p)^{n-1} + np^2(n-1)(1)(q + p)^{n-2}$

$E(X^2) = np(1)^{n-1} + np^2(n-1)(1)^{n-2}$

$E(X^2) = np + np^2(n-1)$

Step 4: Simplify and Compare with Options

We have $E(X^2) = np + np^2(n-1)$. Let's expand and rearrange this expression:

$E(X^2) = np + n^2p^2 - np^2$

$E(X^2) = np(1 - p) + n^2p^2$

Since $q = 1 - p$, we can write:

$E(X^2) = npq + n^2p^2$

Now let's compare this result with the given options:

• Option 1: $nq(np + q) = n^2qp + nq^2$ (Does not match)
• Option 2: $np(p + nq) = np^2 + n^2pq$ (Does not match)
• Option 3: $nq(p + nq) = nqp + n^2q^2$ (Does not match)
• Option 4: $np(np + q) = np \cdot np + np \cdot q = n^2p^2 + npq$ (Matches)

The expression $np(np + q)$ is equivalent to $n^2p^2 + npq$, which matches our calculated value for $E(X^2)$.

Alternatively, for a Binomial distribution B(n, p), we know the mean $E(X) = np$ and the variance $Var(X) = npq$. The variance is related to $E(X)$ and $E(X^2)$ by the formula $Var(X) = E(X^2) - [E(X)]^2$.

Rearranging the formula, we get $E(X^2) = Var(X) + [E(X)]^2$.

Substituting the known values for the Binomial distribution:

$E(X^2) = npq + (np)^2$

$E(X^2) = npq + n^2p^2$

This result is the same as derived from the MGF differentiation method and matches option 4.

Revision Table: MGF and Moments

Additional Information on Moment Generating Functions

The MGF uniquely determines the probability distribution of a random variable. This means if two random variables have the same MGF, they must have the same distribution. MGFs are particularly useful for finding moments and sums of independent random variables.

• The MGF of a sum of independent random variables is the product of their individual MGFs: If $X_1, X_2, \dots, X_k$ are independent, then $M_{X_1 + \dots + X_k}(t) = M_{X_1}(t) \cdot M_{X_2}(t) \cdots M_{X_k}(t)$.
• The MGF of a constant multiplied by a random variable: $M_{aX}(t) = E(e^{t(aX)}) = E(e^{(at)X}) = M_X(at)$.
• The MGF of a constant added to a random variable: $M_{X+c}(t) = E(e^{t(X+c)}) = E(e^{tX}e^{tc}) = e^{tc}E(e^{tX}) = e^{tc}M_X(t)$.

While powerful, MGFs do not always exist for all distributions (e.g., Cauchy distribution). Characteristic functions, on the other hand, always exist.