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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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