Mean and Variance of Random Variable with Given PMF

We are given a non-negative integer valued random variable X with the following probability mass function (PMF):

\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)

We need to find the mean (\(E[X]\)) and variance (\(Var[X]\)) of this random variable X.

Recognizing the Probability Distribution

Let's examine the form of the PMF. The PMF for a Negative Binomial distribution NB(r, p), representing the number of failures before the r-th success in Bernoulli trials with success probability p, is given by:

\( P(Y = y) = \binom{y+r-1}{y} p^r (1-p)^y \quad \text{for } y = 0, 1, 2, ... \)

Let's rewrite the given PMF:

\( P(X=x) = \frac{x+1}{2^{x+2}} = \frac{x+1}{4 \cdot 2^x} = (x+1) \cdot \frac{1}{4} \cdot (\frac{1}{2})^x \)

We can write \( \frac{1}{4} = (\frac{1}{2})^2 \). So the PMF becomes:

\( P(X=x) = (x+1) (\frac{1}{2})^2 (\frac{1}{2})^x \)

Recall that \( \binom{n}{k} = \binom{n}{n-k} \). So, \( \binom{x+1}{x} = \binom{x+1}{(x+1)-x} = \binom{x+1}{1} = x+1 \). Using this, we can write:

\( P(X=x) = \binom{x+1}{x} (\frac{1}{2})^2 (\frac{1}{2})^x \)

Comparing this with the Negative Binomial PMF \( P(Y=y) = \binom{y+r-1}{y} p^r (1-p)^y \), we can see that X follows a Negative Binomial distribution with parameters \( r=2 \) and \( p=1/2 \). Specifically, \( y=x \), \( r=2 \), \( p=1/2 \), and \( 1-p = 1 - 1/2 = 1/2 \). The term \( \binom{x+1}{x} \) matches \( \binom{x+2-1}{x} \).

Calculating the Mean \( E[X] \)

For a Negative Binomial distribution NB(r, p), the mean is given by the formula:

\( E[X] = \frac{r(1-p)}{p} \)

Substituting the identified parameters \( r=2 \) and \( p=1/2 \):

\( E[X] = \frac{2(1 - 1/2)}{1/2} \)

\( E[X] = \frac{2(1/2)}{1/2} \)

\( E[X] = \frac{1}{1/2} \)

\( E[X] = 1 \times 2 = 2 \)

The mean of the random variable X is 2.

Calculating the Variance \( Var[X] \)

For a Negative Binomial distribution NB(r, p), the variance is given by the formula:

\( Var[X] = \frac{r(1-p)}{p^2} \)

Substituting the identified parameters \( r=2 \) and \( p=1/2 \):

\( Var[X] = \frac{2(1 - 1/2)}{(1/2)^2} \)

\( Var[X] = \frac{2(1/2)}{1/4} \)

\( Var[X] = \frac{1}{1/4} \)

\( Var[X] = 1 \times 4 = 4 \)

The variance of the random variable X is 4.

Conclusion

The mean of the random variable X is 2 and the variance of the random variable X is 4.

Revision Table: Probability Distribution Properties

Additional Information: Connections to Negative Binomial Distribution

The Negative Binomial distribution arises in various contexts, often related to sequences of Bernoulli trials.

• Relation to Geometric Distribution: A Negative Binomial distribution NB(r, p) can be thought of as the sum of r independent and identically distributed Geometric(p) random variables, where the Geometric distribution is defined as the number of failures before the first success. The PMF used in this problem is for the number of failures (x) before the r=2 successes. The mean of Geom(p) is \((1-p)/p\) and variance is \((1-p)/p^2\). The sum of r such variables has mean \(r \times (1-p)/p\) and variance \(r \times (1-p)/p^2\), matching the NB formulas.
• Alternative Definition: Some texts define the Negative Binomial distribution as the number of trials until the r-th success. If Y is the number of failures before the r-th success (our definition, x), then the number of trials is Z = Y + r. The PMF for Z is \( P(Z=z) = \binom{z-1}{r-1} p^r (1-p)^{z-r} \) for \( z=r, r+1, \dots \). The mean is \( r/p \) and variance is \( r(1-p)/p^2 \). It's important to check the definition being used. In this problem, X is the number of failures (starts at x=0), matching our definition of NB(r, p) with mean \( r(1-p)/p \).
• Applications: Negative Binomial distribution is used to model phenomena like the number of defective items found before a certain number of non-defective items are found, or the number of customers arriving before the N-th customer who makes a purchase.