On Main Street, a driver’s speed just before an accident is uniformly distributed on [5, 20]. Given the speed, the resulting loss from the accident is exponentially distributed with mean equal to three times the speed.

Calculate the variance of a loss due to an accident on Main Street.

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

Step 1: Define Random Variables and Conditional Distributions

Let [math]S[/math] be the random variable representing the driver's speed, and [math]X[/math] be the random variable representing the resulting loss from the accident. The problem provides the following information:

• Speed Distribution: The driver's speed [math]S[/math] is uniformly distributed on the interval [math][5, 20][/math].

The probability density function (PDF) for [math]S[/math] is:

[[math]]f_S(s) = \frac{1}{20 - 5} = \frac{1}{15} \quad \text{for } 5 \le s \le 20[[/math]]

• Conditional Loss Distribution: Given the speed [math]S=s[/math], the resulting loss [math]X[/math] is exponentially distributed with a mean equal to three times the speed.

Therefore, the conditional mean of [math]X[/math] given [math]S[/math] is:

[[math]]\operatorname{E}[X | S] = 3S[[/math]]

For an exponentially distributed random variable with mean [math]\mu[/math], the variance is [math]\mu^2[/math]. Thus, the conditional variance of [math]X[/math] given [math]S[/math] is:

[[math]]\operatorname{Var}(X | S) = (3S)^2 = 9S^2[[/math]]

Step 2: Apply the Law of Total Variance

To calculate the variance of the loss [math]\operatorname{Var}(X)[/math], we use the Law of Total Variance, which states:

[[math]]\operatorname{Var}(X) = \operatorname{E}[\operatorname{Var}(X | S)] + \operatorname{Var}[\operatorname{E}(X | S)][[/math]]

Substituting the expressions for [math]\operatorname{E}[X | S][/math] and [math]\operatorname{Var}(X | S)[/math] from Step 1:

[[math]]\operatorname{Var}(X) = \operatorname{E}[9S^2] + \operatorname{Var}[3S][[/math]]

We can simplify the terms:

• [math]\operatorname{E}[9S^2] = 9\operatorname{E}[S^2][/math]
• [math]\operatorname{Var}[3S] = 3^2 \operatorname{Var}(S) = 9\operatorname{Var}(S)[/math]

So, the equation becomes:

[[math]]\operatorname{Var}(X) = 9\operatorname{E}[S^2] + 9\operatorname{Var}(S)[[/math]]

Step 3: Calculate Moments of the Speed Distribution

The speed [math]S[/math] is uniformly distributed on [math][a, b] = [5, 20][/math]. We need to find [math]\operatorname{E}[S^2][/math] and [math]\operatorname{Var}(S)[/math].

• Mean of S:

For a uniform distribution on [math][a, b][/math], the mean is [math]\operatorname{E}[S] = \frac{a+b}{2}[/math].

[[math]]\operatorname{E}[S] = \frac{5+20}{2} = \frac{25}{2} = 12.5[[/math]]

• Variance of S:

For a uniform distribution on [math][a, b][/math], the variance is [math]\operatorname{Var}(S) = \frac{(b-a)^2}{12}[/math].

[[math]]\operatorname{Var}(S) = \frac{(20-5)^2}{12} = \frac{15^2}{12} = \frac{225}{12} = 18.75[[/math]]

• Second Moment of S:

The second moment [math]\operatorname{E}[S^2][/math] can be calculated using the relationship [math]\operatorname{Var}(S) = \operatorname{E}[S^2] - (\operatorname{E}[S])^2[/math], so [math]\operatorname{E}[S^2] = \operatorname{Var}(S) + (\operatorname{E}[S])^2[/math].

[[math]]\operatorname{E}[S^2] = 18.75 + (12.5)^2 = 18.75 + 156.25 = 175[[/math]]

Step 4: Compute the Variance of the Loss

Now we substitute the calculated values of [math]\operatorname{E}[S^2][/math] and [math]\operatorname{Var}(S)[/math] into the equation for [math]\operatorname{Var}(X)[/math] from Step 2:

[[math]]\operatorname{Var}(X) = 9\operatorname{E}[S^2] + 9\operatorname{Var}(S)[[/math]]

[[math]]\operatorname{Var}(X) = 9(175) + 9(18.75)[[/math]]

[[math]]\operatorname{Var}(X) = 1575 + 168.75[[/math]]

[[math]]\operatorname{Var}(X) = 1743.75[[/math]]

The variance of a loss due to an accident on Main Street is [math]1743.75[/math].

Key Insights

• The Law of Total Variance is a fundamental tool for calculating the variance of a random variable when its distribution depends on another random variable. It states [math]\operatorname{Var}(X) = \operatorname{E}[\operatorname{Var}(X | Y)] + \operatorname{Var}[\operatorname{E}(X | Y)][/math].
• Understanding the properties of common distributions is crucial. For an exponential distribution with mean [math]\mu[/math], its variance is [math]\mu^2[/math].
• For a uniform distribution on [math][a, b][/math], the mean is [math]\operatorname{E}[S] = \frac{a+b}{2}[/math] and the variance is [math]\operatorname{Var}(S) = \frac{(b-a)^2}{12}[/math].
• The second moment of a random variable [math]Y[/math], [math]\operatorname{E}[Y^2][/math], is related to its mean and variance by [math]\operatorname{E}[Y^2] = \operatorname{Var}(Y) + (\operatorname{E}[Y])^2[/math]. This relationship is often used to find [math]\operatorname{E}[Y^2][/math] when [math]\operatorname{E}[Y][/math] and [math]\operatorname{Var}(Y)[/math] are known.
• Constants can be factored out of expectation and variance operators: [math]\operatorname{E}[cY] = c\operatorname{E}[Y][/math] and [math]\operatorname{Var}(cY) = c^2\operatorname{Var}(Y)[/math].

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Solution: E

Let S be the speed and X be the loss. Given S, X has an exponential distribution with mean 3X. Then, noting that the variance of an exponential random variable is the square of the mean, the variance of a uniform random variable is the square of the range divided by 12, and for any random variable the second moment is the variance plus the square of the mean:

[[math]] \begin{align*} \operatorname{Var}(X) &= \operatorname{Var}[\operatorname{E}(X | S) ] + \operatorname{E}[ \operatorname{Var}(X | S) ] \\ &= \operatorname{Var}[3S] + \operatorname{E}[9S^2] \\ &= 9(20-5)^2/12 + 9[(20-5)^2/12 + 12.5^2] \\ &= 1743.75 \end{align*} [[/math]]

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