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Calculate moments for linear combinations of independent random variables

Calculate moments for linear combinations of independent random variables

In the previous reading, we defined \(Y=c_{1} X_{1}+c_{2} X_{2}+\cdots+c_{p} X_{p}\) to be a linear combination of the independent random variables \(X_{1}, X_{2}, \ldots, X_{p}\) where \(c_{1}, c_{2}, \ldots, c_{p}\) are constants.

Now, we may wish to calculate moments such as mean and variance for these linear combinations.

Mean of a Linear Combination of Independent Random Variables

Let \(Y=c_{1} X_{1}+c_{2} X_{2}+\cdots+c_{p} X_{p}\) be a linear combination of random variables \(X_{1}, X_{2}, \ldots, X_{p}\). Assume that \(X_{1}, X_{2}, \ldots, X_{p}\) are independent. Then, the expected value of \(Y\) is given by:
$$
\text{E}(\text{Y})=\text{c}_{1} \text{E}\left(\text{X}_{1}\right)+\text{c}_{2} \text{E}\left(\text{X}_{2}\right)+\cdots+\text{c}_{\text{p}} \text{E}\left(\text{X}_{\text{p}}\right)
$$

Example 1: Mean of a Linear Combination of Independent Random Variables

An insurance company insures farmers against losses caused to crops by hail. Let \(\text{X}, \text{Y}\), and \(\text{Z}\) be the amount of claims from three different farmers. The mean claim amounts are 20, 15, and 25 for \(X, Y\), and Z, respectively.

Assume that \(\text{X}, \text{Y}\), and \(\text{Z}\) are independent.

Find \(E(X+Y+Z)\)

Solution

We know that,

$$
\begin{aligned}
\text{E}(\text{X}+\text{Y}+\text{Z}) &=\text{E}(\text{X})+\text{E}(\text{Y})+\text{E}(\text{Z}) \\
&=20+15+25=60 \end{aligned}
$$

Example 2: Mean of a Linear Combination of Independent Random Variables

A sickness benefit plan has two types of benefits, \(X\) and \(Y\). The benefits are independent and with the following probability density functions:
$$
\begin{aligned}
&\text{f(x})=\left\{\begin{array}{cc}
0.2 e^{-0.2\text{x}}, \quad \text{x} \geq 0 \\
0, \quad \text { elsewhere }
\end{array}\right. \\ \\
&\text{f(y)}=\left\{\begin{array}{cc}
0.1 e^{-0.1 \text{y}}, \quad \text{y} \geq 0 \\
0, \quad \text { elsewhere }
\end{array}\right.
\end{aligned}
$$

Find \(\text{E}(\text{X}-\text{Y})\)

Solution

Since \(\text{X}\) and \(\text{Y}\) are independent, then
$$
\text{E}(\text{X}-\text{Y})=\text{E}(\text{X})-\text{E}(\text{Y})
$$
Now,
$$
\text{E}(\text{X})=\int_{0}^{\infty} \text{x} * 0.2 \text{e}^{-0.2 \text{x}} \text{dx}=5
$$
and,
$$
\begin{gathered}
\text{E}(\text{Y})=\int_{0}^{\infty} \text{y} * 0.1 \text{e}^{-0.1 \text{y}} \text{dy}=10 \\ \\
\Rightarrow \text{E}(\text{X}-\text{Y})=5-10=-5
\end{gathered}
$$

Variance of a Linear Combination of Independent Random Variables

Let \(Y=c_{1} X_{1}+c_{2} X_{2}+\cdots+c_{p} X_{p}\) be a linear combination of random variables \(X_{1}, X_{2}, \ldots, X_{p}\). Assume that \(X_{1}, X_{2}, \ldots, X_{p}\) are independent. Then, the variance of \(Y\) is given by:

$$
\text{V}(\text{Y})=\text{c}_{1}^{2} \text{V}\left(\text{X}_{1}\right)+\text{c}_{2}^{2} \text{V}\left(\text{X}_{2}\right)+\cdots+\text{c}_{\text{p}}^{2} \text{V}\left(\text{X}_{\text{p}}\right)
$$

Example 1: Variance of a Linear Combination of Independent Random Variables

An actuary wants to determine the annual number of hailstorm days in two neighboring counties, \(\text{P}\) and Q. Let \(X\) and \(Y\) be the annual number of hailstorm days in counties \(P\) and \(Q\), respectively. \(X\) and \(Y\) are independent and distributed as in the table below:

$$\begin{array}{l|c|c|c|c}
& 0 & 1 & 2 & 3 \\
\hline \text{P}(\text{x})& \frac{1}{3}& \frac{1}{6}& \frac{1}{3} & \frac{1}{6}\\
\hline \text{P}(\text{y})& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}\\ \end{array}$$

Assume that \(\text{X}\) and \(\text{Y}\) are independent.

Find \(\operatorname{Var}(\text{X}-\text{Y})\)

Solution

Since \(\text{X}\) and \(\text{Y}\) are independent,
$$
\operatorname{Var}(\text{X}-\text{Y})=\operatorname{Var}(\text{X})+\operatorname{Var}(\text{Y})
$$

We know that,

$$
\begin{aligned}
\operatorname{Var}(\text{X}) &=\text{E}\left(\text{X}^{2}\right)-[\text{E}(\text{X})]^{2} \\
&=0^{2} * \frac{1}{3}+1^{2} * \frac{1}{6}+2^{2} * \frac{1}{3}+3^{2} * \frac{1}{6}-\left(0 * \frac{1}{3}+1 * \frac{1}{6}+2 * \frac{1}{3}+3 * \frac{1}{6}\right)^{2} \\
&=\frac{11}{9} \\ \\
\operatorname{Var}(\text{Y}) &=\text{E}\left(\text{Y}^{2}\right)-[\text{E}(\text{Y})]^{2} \\
&=0^{2} * \frac{1}{4}+1^{2} * \frac{1}{4}+2^{2} * \frac{1}{4}+3^{2} * \frac{1}{4}-\left(0 * \frac{1}{4}+1 * \frac{1}{4}+2 * \frac{1}{4}+3 * \frac{1}{4}\right)^{2} \\
&=\frac{5}{4}
\end{aligned}
$$

Therefore,
$$
\operatorname{Var}(\text{X}-\text{Y})=\frac{11}{9}+\frac{5}{4}=2.4722
$$

Example 2: Variance of a Linear Combination of Independent Random Variables

Let \(\text{X}\) and \(\text{Y}\) be the random variables for the amount of whole life insurance benefit and term insurance benefit. \(\text{X}\) and \(\text{Y}\) are independent and with the following probability density functions:

$$
\begin{aligned}
&f(x)=\left\{\begin{array}{c}
0.02 e^{-0.02 x}, \quad x \geq 0 \\
0, \quad \text { elsewhere }
\end{array}\right. \\ \\
&f(y)=\left\{\begin{array}{cc}
0.01 e^{-0.01 y}, \quad y \geq 0 \\
0, \quad \text { elsewhere }
\end{array}\right.
\end{aligned}
$$

Find \(\operatorname{Var}(X+Y)\)

Solution

Since \(\text{X}\) and \(\text{Y}\) are independent, then

$$
\operatorname{Var}(\text{X}+\text{Y})=\operatorname{Var}(\text{X})+\operatorname{Var}(\text{Y})
$$

Now,
$$
\begin{aligned}
\operatorname{Var}(\text{X}) &=\text{E}\left(\text{X}^{2}\right)-[\text{E}(\text{X})]^{2} \\
&=\int_{0}^{\infty} \text{x}^{2} * 0.02 \text{e}^{-0.02 \text{x}} \text{dx}-\left(\int_{0}^{\infty} \text{x} * 0.02 \text{e}^{-0.02 \text{x}} \text{dx}\right)^{2} \\
&=2,500
\end{aligned}
$$

and,
$$
\begin{aligned}
\operatorname{Var}(\text{Y})&=\text{E}\left(\text{Y}^{2}\right)-[\text{E}(\text{Y})]^{2}\\&=\int_{0}^{\infty} \mathrm{y}^{2} * 0.01 \mathrm{e}^{-0.01 \mathrm{y}} \mathrm{dy}-\left(\int_{0}^{\infty} \mathrm{y} * 0.01 \mathrm{e}^{-0.01 \mathrm{y}} \mathrm{dy}\right)^{2} \\&=10,000
\end{aligned}
$$

Therefore,
$$
\operatorname{Var}(\text{X}+\text{Y})=2,500+10,000=12,500
$$

 

Learning Outcome

Topic 3. h: Multivariate Random Variables – Calculate moments for linear combinations of independent random variables.

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