###### Calculate joint moments, such as the c ...

Let \(\text{X}\) and \(\text{Y}\) be two discrete random variables, with a joint probability... **Read More**

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.

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)

$$

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

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}

$$

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

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}

$$

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)

$$

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

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

$$

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

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.**