Suppose that in a game, killing a certain monster has a fixed "drop rate" for equipment 1 and 2. Assume that after killing the monster, equipment 1 drops with probability , equipment 2 drops with probability , and no equipment drops with probability . Let denote the minimum number of kills needed to collect both equipment 1 and 2. Find the expectation and variance of .

Stochastic Processes Solution – Problem 44: Find the expectation and variance of

Question

Suppose that in a game, killing a certain monster has a fixed "drop rate" for equipment 1 and 2. Assume that after killing the monster, equipment 1 drops with probability $0.2$ , equipment 2 drops with probability $0.1$ , and no equipment drops with probability $0.7$ . Let $\tau$ denote the minimum number of kills needed to collect both equipment 1 and 2. Find the expectation and variance of $\tau$ .

Step-by-step solution

Each kill has three mutually exclusive and exhaustive outcomes: drop equipment 1 (probability $p=0.2$ ), drop equipment 2 (probability $q=0.1$ ), or drop nothing (probability $r=0.7$ ), and the kills are mutually independent. Define $X$ as the number of kills needed to first obtain either equipment 1 or equipment 2, and $Y$ as the number of additional kills needed after obtaining one piece of equipment to obtain the other. Clearly $\tau = X + Y$ , and $X$ and $Y$ are independent, so $E[\tau] = E[X] + E[Y]$ and $\text{Var}(\tau) = \text{Var}(X) + \text{Var}(Y)$ .

Analysis of the distribution of $X$ : $X$ is the number of kills until the first "success," where "success" means dropping equipment 1 or equipment 2, with success probability $s = p + q = 0.3$ . $X$ follows a geometric distribution with parameter $s=0.3$ (defined as $P(X=k) = (1-s)^{k-1}s$ , $k=1,2,\dots$ ), with expectation $E[X] = 1/s$ and variance $\text{Var}(X) = (1-s)/s^2$ . Substituting $s=0.3$ : $E[X] = 10/3$ , $\text{Var}(X) = 70/9$ .

Analysis of $E[Y]$ : Use the law of total expectation. Let $Z$ denote the type of equipment obtained first ( $Z=1$ for equipment 1, $Z=2$ for equipment 2). Then $P(Z=1) = p/s = 2/3$ , $P(Z=2) = q/s = 1/3$ . When $Z=1$ , we need to first obtain equipment 2, with per-trial success probability $q=0.1$ , giving geometric expectation $1/q$ ; when $Z=2$ , we need equipment 1 with per-trial success probability $p=0.2$ , giving expectation $1/p$ . Thus $E[Y|Z=1] = 10$ , $E[Y|Z=2] = 5$ , and by the law of total expectation: $E[Y] = 10 \times 2/3 + 5 \times 1/3 = 25/3$ .

Analysis of $\text{Var}(Y)$ : Use the law of total variance $\text{Var}(Y) = E[\text{Var}(Y|Z)] + \text{Var}(E[Y|Z])$ . First compute the expectation of the conditional variance: $\text{Var}(Y|Z=1) = (1-q)/q^2 = 90$ , $\text{Var}(Y|Z=2) = (1-p)/p^2 = 20$ , so $E[\text{Var}(Y|Z)] = 90 \times 2/3 + 20 \times 1/3 = 200/3$ . Next compute the variance of $E[Y|Z]$ : $E[Y|Z]$ takes values 10 and 5 with probabilities $2/3$ and $1/3$ respectively, so $E[(E[Y|Z])^2] = 100 \times 2/3 + 25 \times 1/3 = 75$ , and $(E[Y])^2 = (25/3)^2 = 625/9$ , giving $\text{Var}(E[Y|Z]) = 75 - 625/9 = 50/9$ . Therefore $\text{Var}(Y) = 200/3 + 50/9 = 650/9$ .

$E[\tau] = 10/3 + 25/3 = 35/3$ , $\text{Var}(\tau) = 70/9 + 650/9 = 720/9 = 80$ .

Final answer

Expectation: $\dfrac{35}{3}$ , Variance: $80$

Marking scheme

The following is the rubric based on the official solution.

1. Checkpoints (max 7 pts)

Select any one logical chain for scoring | take the maximum among chains; do not accumulate across chains.

Chain A: Random Variable Decomposition ( $\tau = X + Y$ , Official Solution)

First phase $X$ (first obtaining any equipment)
Correctly identify that $X$ follows a geometric distribution with parameter $p=0.3$ and obtain $E[X] = 10/3$ . [1 pt]
Correctly compute the variance $\text{Var}(X) = 70/9$ . [1 pt]
Second phase $Y$ (obtaining the remaining equipment) -- expectation
Correctly write the conditional probabilities/weights for entering the second phase: probability $2/3$ ( $p_1/s$ ) of first obtaining equipment 1, probability $1/3$ ( $p_2/s$ ) of first obtaining equipment 2. [1 pt]
Use the law of total expectation to compute $E[Y] = 25/3$ , and hence the total expectation $E[\tau] = 35/3$ . [1 pt]
Second phase $Y$ -- variance (core difficulty)
Correctly compute the conditional variances (90 and 20 respectively) or conditional second moments (190 and 45 respectively). [1 pt]
Correctly obtain $\text{Var}(Y) = 650/9$ .
*Scoring criterion: Must use the law of total variance (including the $\text{Var}(E[Y|Z])$ term) or compute via $E[Y^2] - (E[Y])^2$ . If only the weighted average of conditional variances $\sum p_i \text{Var}_i$ is computed, this point is not awarded.* [1 pt]
Final result
Use independence $\text{Var}(\tau) = \text{Var}(X) + \text{Var}(Y)$ to obtain the correct answer $80$ . [1 pt]

Chain B: Markov Chain / System of Linear Equations

Expectation equations
Correctly set up the system of linear equations for the expected number of steps from each state (e.g., $E_0, E_1, E_2$ ). [2 pts]
Solve the system to obtain the correct total expectation $35/3$ . [1 pt]
Variance / second moment equations
Correctly set up the system of equations for the second moments ( $E[\tau^2]$ ) or variances from each state. [2 pts]
Solve for the key second moment values or intermediate variables. [1 pt]
Final result
Correctly compute $\text{Var}(\tau) = E[\tau^2] - (E[\tau])^2 = 80$ . [1 pt]

2. Zero-credit items

Merely listing the geometric distribution formula (e.g., $E=1/p$ ) without performing specific calculations with the probabilities $0.2, 0.1, 0.7$ from this problem.
Incorrectly adding expectations: $E[\tau] = 1/0.2 + 1/0.1$ (ignoring mutual exclusivity and the probability of no drop).
Merely copying the probability values from the problem with no derivation.

3. Deductions

Computational error: Each obvious arithmetic error (not a logical error), deduct 1 pt.
Law of total variance misuse (logical flaw): When computing $\text{Var}(Y)$ , if the student only computes the weighted average of conditional variances ( $E[\text{Var}(Y|Z)]$ ) and omits the "variance of the conditional expectation" term ( $\text{Var}(E[Y|Z])$ ), yielding $\text{Var}(Y)=600/9$ or similar. This is a major logical gap; the step earns no credit (already reflected in Checkpoints; if there is score overflow, deduct 1 pt, but the minimum is 0).
Independence justification missing: Directly adding the variances of $X$ and $Y$ without mentioning independence (or the Markov property), but with correct calculations. Given the undergraduate level, no deduction.

Stochastic Processes – Problem 44: Find the expectation and variance of