Key Concepts and Formulas

• Discrete Probability Distribution Validity: A function $P(X=x)$ represents a valid discrete probability distribution if:
  1. $P(X=x) \ge 0$ for all possible values of $X$.
  2. $\sum_x P(X=x) = 1$.
• Mean (Expected Value) of a Discrete Random Variable: For a discrete random variable $X$ with possible values $x_i$ and probabilities $P(X=x_i)$, the mean $E[X]$ is given by: $E[X] = \sum_{x_i} x_i \cdot P(X=x_i)$
• Sum of an Infinite Geometric Series: The sum of an infinite geometric series $a + ar + ar^2 + \dots$ is $S = \frac{a}{1-r}$, provided $|r| < 1$.
• Sum of an Arithmetic-Geometric Series: The sum of the series $\sum_{j=1}^{\infty} j x^j$ for $|x|<1$ is given by $\frac{x}{(1-x)^2}$.

Step-by-Step Solution

First, let's verify the given probability function represents a valid distribution.

Step 1: Verify the Probability Distribution We need to check two conditions:

1. All probabilities $P(X=x)$ are non-negative.
2. The sum of all probabilities $\sum_{x} P(X=x)$ equals 1.

• Non-negativity:

  • $P(X=0) = \frac{1}{2} \ge 0$.
  • For $j \ge 1$, $P(X=j) = \frac{1}{3^j}$ is always positive, thus $\ge 0$. Condition 1 is satisfied.

• Sum of probabilities: $\sum_{x} P(X=x) = P(X=0) + \sum_{j=1}^{\infty} P(X=j)$ Substitute the given probability functions: $= \frac{1}{2} + \sum_{j=1}^{\infty} \left(\frac{1}{3^j}\right)$ The sum $\sum_{j=1}^{\infty} \left(\frac{1}{3}\right)^j$ is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$. Since $|r| < 1$, the series converges. $\sum_{j=1}^{\infty} \left(\frac{1}{3}\right)^j = \frac{a}{1-r} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$ Now, substitute this back into the total probability sum: $\sum_{x} P(X=x) = \frac{1}{2} + \frac{1}{2} = 1$ Condition 2 is satisfied. The given function is a valid probability distribution.

Part 1: Calculate the Mean of the Distribution ($E[X]$)

Step 2: Set up the formula for the mean The mean $E[X]$ is calculated as $\sum_{x} x \cdot P(X=x)$. Since the probability function is defined in two parts ($X=0$ and $X=j$ for $j \ge 1$), we split the summation accordingly. $E[X] = (0 \cdot P(X=0)) + \sum_{j=1}^{\infty} (j \cdot P(X=j))$

Step 3: Substitute the probabilities and simplify Substitute $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$: $E[X] = \left(0 \cdot \frac{1}{2}\right) + \sum_{j=1}^{\infty} \left(j \cdot \frac{1}{3^j}\right)$ The first term $0 \cdot \frac{1}{2}$ is $0$. $E[X] = \sum_{j=1}^{\infty} j \left(\frac{1}{3}\right)^j$

Step 4: Evaluate the arithmetic-geometric series This is an arithmetic-geometric series of the form $\sum_{j=1}^{\infty} j x^j$. Using the formula $\sum_{j=1}^{\infty} j x^j = \frac{x}{(1-x)^2}$ with $x = \frac{1}{3}$: $E[X] = \frac{1/3}{(1 - 1/3)^2}$ $E[X] = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9}$ To simplify, multiply by the reciprocal: $E[X] = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$ The mean of the distribution is $\frac{3}{4}$.

Part 2: Calculate P(X is positive and even)

Step 5: Identify the values of X that satisfy the condition "X is positive and even" means $X$ can take values $2, 4, 6, 8, \dots$.

Step 6: Set up the summation for the probability $P(X \text{ is positive and even}) = P(X=2) + P(X=4) + P(X=6) + \dots$

Step 7: Substitute the probabilities for these values Since $j \ge 1$, we use the formula $P(X=j) = \frac{1}{3^j}$: $P(X \text{ is positive and even}) = \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \dots$

Step 8: Evaluate the infinite geometric series This is an infinite geometric series.

• The first term $a = \frac{1}{3^2} = \frac{1}{9}$.
• The common ratio $r = \frac{1/3^4}{1/3^2} = \frac{1}{3^2} = \frac{1}{9}$. Since $|r| < 1$, the series converges. Using the formula $S = \frac{a}{1-r}$: $P(X \text{ is positive and even}) = \frac{1/9}{1 - 1/9} = \frac{1/9}{8/9}$ $P(X \text{ is positive and even}) = \frac{1}{8}$ The probability that $X$ is positive and even is $\frac{1}{8}$.

Common Mistakes & Tips

• Checking Distribution Validity: Always perform the validity check for a probability distribution. It ensures you haven't misunderstood the problem or that the problem statement itself is consistent.
• Splitting Summations: Pay close attention to the definition of the probability function, especially when it's defined piecewise (e.g., differently for $X=0$ and $X \ge 1$). Incorrectly applying a general formula to a specific case (like $P(X=0)$) is a common error.
• Identifying Series Correctly: Accurately identify the first term ($a$) and common ratio ($r$) for geometric series. For sums involving only even (or odd) terms, the common ratio will be $r^2$ (e.g., $(1/3)^2 = 1/9$) if the original ratio was $r$.

Summary

We first verified that the given probability function constitutes a valid probability distribution. Then, we calculated the mean (expected value) of the distribution by summing $x \cdot P(X=x)$ over all possible values of $X$, which involved evaluating an arithmetic-geometric series. The mean was found to be $\frac{3}{4}$. Finally, we calculated the probability that $X$ is positive and even by summing $P(X=j)$ for $j=2, 4, 6, \dots$, which involved evaluating another infinite geometric series. This probability was found to be $\frac{1}{8}$.

The final answer is $\boxed{\frac{3}{4} \text{ and } \frac{1}{8}}$, which corresponds to option (B).