Let [math]N[/math] denote the number of accidents occurring during one month on the northbound side of a highway and let [math]S[/math] denote the number occurring on the southbound side. Suppose that [math]N[/math] and [math]S[/math] are jointly distributed as indicated in the table.

N\S	0	1	2	3 or more
0	0.04	0.06	0.10	0.04
1	0.10	0.18	0.08	0.03
2	0.12	0.06	0.05	0.02
3 or more	0.05	0.04	0.02	0.01

Calculate [math]\operatorname{Var}(N | N + S = 2) [/math].

• 0.48
• 0.55
• 0.67
• 0.91
• 1.25

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Identify the Conditional Event and Possible Outcomes

The question asks to calculate the variance of [math]N[/math] given the condition [math]N + S = 2[/math]. First, we need to identify all possible pairs of [math](N, S)[/math] from the given joint distribution table that satisfy the condition [math]N + S = 2[/math] and their associated joint probabilities. The given joint probability table is:

N\S	0	1	2	3 or more
0	0.04	0.06	0.10	0.04
1	0.10	0.18	0.08	0.03
2	0.12	0.06	0.05	0.02
3 or more	0.05	0.04	0.02	0.01

The pairs [math](N, S)[/math] for which [math]N + S = 2[/math] are:

• [math](N=0, S=2)[/math]: From the table, [math]P(N=0, S=2) = 0.10[/math].
• [math](N=1, S=1)[/math]: From the table, [math]P(N=1, S=1) = 0.18[/math].
• [math](N=2, S=0)[/math]: From the table, [math]P(N=2, S=0) = 0.12[/math].

These are the only possible outcomes from the table where [math]N+S=2[/math].

Step 2: Calculate the Probability of the Conditional Event

The probability of the conditioning event [math]N+S=2[/math] is the sum of the joint probabilities of the outcomes identified in Step 1:

[[math]]P(N+S=2) = P(N=0, S=2) + P(N=1, S=1) + P(N=2, S=0)[[/math]]

[[math]]P(N+S=2) = 0.10 + 0.18 + 0.12[[/math]]

[[math]]P(N+S=2) = 0.40[[/math]]

Step 3: Calculate Conditional Probabilities for N

Now we calculate the conditional probability mass function for [math]N[/math] given [math]N+S=2[/math]. The general formula is [math]P(N=n | N+S=2) = \frac{P(N=n, S=2-n)}{P(N+S=2)}[/math].

• For [math]N=0[/math]:

[[math]]P(N=0 | N+S=2) = \frac{P(N=0, S=2)}{P(N+S=2)} = \frac{0.10}{0.40} = 0.25[[/math]]

• For [math]N=1[/math]:

[[math]]P(N=1 | N+S=2) = \frac{P(N=1, S=1)}{P(N+S=2)} = \frac{0.18}{0.40} = 0.45[[/math]]

• For [math]N=2[/math]:

[[math]]P(N=2 | N+S=2) = \frac{P(N=2, S=0)}{P(N+S=2)} = \frac{0.12}{0.40} = 0.30[[/math]]

These conditional probabilities sum to [math]0.25 + 0.45 + 0.30 = 1.00[/math], as expected. We can summarize the conditional distribution of [math]N[/math] given [math]N+S=2[/math] in a table:

[math]N[/math]	[math]P(N=n | N+S=2)[/math]
0	0.25
1	0.45
2	0.30

Step 4: Calculate the Conditional Expectation (Mean) of N

The conditional expectation of [math]N[/math] given [math]N+S=2[/math], denoted as [math]E[N | N+S=2][/math], is calculated using the conditional probabilities from Step 3:

[[math]]E[N | N+S=2] = \sum_{n} n \cdot P(N=n | N+S=2)[[/math]]

[[math]]E[N | N+S=2] = (0 \times 0.25) + (1 \times 0.45) + (2 \times 0.30)[[/math]]

[[math]]E[N | N+S=2] = 0 + 0.45 + 0.60[[/math]]

[[math]]E[N | N+S=2] = 1.05[[/math]]

Step 5: Calculate the Conditional Second Moment of N

The conditional second moment of [math]N[/math] given [math]N+S=2[/math], denoted as [math]E[N^2 | N+S=2][/math], is calculated as:

[[math]]E[N^2 | N+S=2] = \sum_{n} n^2 \cdot P(N=n | N+S=2)[[/math]]

[[math]]E[N^2 | N+S=2] = (0^2 \times 0.25) + (1^2 \times 0.45) + (2^2 \times 0.30)[[/math]]

[[math]]E[N^2 | N+S=2] = (0 \times 0.25) + (1 \times 0.45) + (4 \times 0.30)[[/math]]

[[math]]E[N^2 | N+S=2] = 0 + 0.45 + 1.20[[/math]]

[[math]]E[N^2 | N+S=2] = 1.65[[/math]]

Step 6: Calculate the Conditional Variance of N

Finally, we can calculate the conditional variance of [math]N[/math] given [math]N+S=2[/math] using the formula:

[[math]]Var(N | N+S=2) = E[N^2 | N+S=2] - (E[N | N+S=2])^2[[/math]]

Substituting the values calculated in Step 4 and Step 5:

[[math]]Var(N | N+S=2) = 1.65 - (1.05)^2[[/math]]

[[math]]Var(N | N+S=2) = 1.65 - 1.1025[[/math]]

[[math]]Var(N | N+S=2) = 0.5475[[/math]]

Rounding to two decimal places, the variance is 0.55.

Key Insights

• To find the conditional distribution of a variable given a sum, identify all joint outcomes that satisfy the sum.
• The probability of the conditioning event is the sum of the joint probabilities of these outcomes.
• Conditional probabilities are calculated by dividing the joint probability of an outcome by the probability of the conditioning event.
• Conditional expectation and variance are computed using the conditional probability mass function, following the standard formulas [math]E[X] = \sum x P(X=x)[/math] and [math]Var(X) = E[X^2] - (E[X])^2[/math].

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Solution: B

Given N + S = 2, there are 3 possibilities (N,S) = (2,0), (1,1), (0,2) with probabilities 0.12, 0.18, and 0.10 respectively. The associated conditional probabilities are

[[math]] \begin{align*} P(N = 0 | N + S = 2) = 0.10/0.40 &= 0.25, \\ P(N = 1 | N + S = 2) = 0.18/0.40 &= 0.45, \\ P(N = 2 | N + S = 2) = 0.12/0.40 &= 0.30. \\ \end{align*} [[/math]]

The mean is 0.25(0) + 0.45(1) + 0.30(2) = 1.05. The second moment is 0.25(0) + 0.45(1) + 0.30(4) = 1.65. The variance is 1.65 – (1.05)(1.05) = 0.5475.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.