###### Stress-Testing

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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be random variables and let \(c_{1}, c_{2}, \ldots, c_{n}\) be constants.

Then,

$$

\text{Y}=\text{c}_{1}

\text{X}_{1}+\text{c}_{2} \text{X}_{2}+\cdots+\text{c}_{\text{n}} \text{X}_{\text{n}}

$$

is a linear combination of \(X_{1}, X_{2}, \ldots, X_{n}\).

In this reading, however, we will only base our discussion on the linear combinations of independent normal random variables.

In statistics, it is usually assumed that a sample is drawn from a population that is normally distributed with mean \(\mu\) and variance \(\sigma^{2}\).

Now, let \(\text{X}_{1}, \text{X}_{2}, \ldots, \text{X}_{\text{n}}\) be independent normal random variables with means \(\mu_{1}, \mu_{2}, \ldots, \mu_{\text{n}}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}, \ldots, \sigma_{\text{n}}^{2}\), respectively. Then, it can be proven that \(\text{X}_{1}+\text{X}_{2}+\cdots+\text{X}_{\text{n}}\) is normally distributed with mean \(\mu_{1}+\mu_{2}+\cdots+\mu_{\text{n}}\) and variance \(\sigma_{1}^{2}+\sigma_{2}^{2}+\ldots+\sigma_{\text{n}}^{2}\).

We can calculate the probability for linear combinations of independent normal random variables. This is illustrated in the following examples:

The annual incomes of three businessmen are normal with means 1000, 2000 and 3000, respectively. The variances of income are 750, 950, and 800, respectively.

Suppose that the annual incomes are independent.

Find the probability that the total annual income is exactly 6,100.

Let \(X_{1}, X_{2}, X_{3}\) be the random variables for the income of the three businessmen, respectively.

Since, \(X_{1}, X_{2}, X_{3}\) are normal with means 1000,2000 and 3000 and variances 750,950 and 800 , respectively, then \(X_{1}+X_{2}+X_{3}\) is normal with mean \(1,000+2,000+3,000=6,000\) and variance \(750+950+800=2,500\)

We wish to find,

$$

\text{P}\left(\text{X}_{1}+\text{X}_{2}+\text{X}_{3}=3,000\right)=\text{P}\left(\text{Z}=\frac{6,100-6,000}{\sqrt{2,500}}\right)=\text{P}(\text{Z}=2)

$$

and from the standard normal table,

$$

\text{P}(\text{Z}=2)=0.9772

$$

Let \(X_{1}\) and \(X_{2}\) be the random variables for the amount of sickness benefits given to two individuals, \(A\) and \(\text{B}\), respectively. \(\text{X}_{1}\) and \(\text{X}_{2}\) are independent with the following probability density functions:

$$

\text{f}\left(\text{x}_{1}\right)=\left\{\begin{array}{c}

\frac{1}{\sqrt{32 \pi}} e^{-\frac{1}{2}\left(\frac{\text{x}_{1}-20}{4}\right)^{2}}, \quad\text{x}_{1} \geq 0 \\

0, \quad \quad \quad \quad \quad\quad\text { elsewhere }

\end{array}\right.

$$

and,

$$

\text{f}\left(\text{x}_{2}\right)=\left\{\begin{array}{c}

\frac{1}{\sqrt{18 \pi}} e^{-\frac{1}{2}\left(\frac{\text{x}_{2}-20}{3}\right)^{2}}, \quad x_{2} \geq 0 \\

0, \quad \text { elsewhere }

\end{array}\right.

$$

Find \(\text{P}\left(\text{X}_{1}+\text{X}_{2} \leq 45\right)\)

From the given pdfs, we can see that \(X_{1}\) and \(X_{2}\) are normal random variables with means 20 and 30, respectively, and variances 16 and 9, respectively.

Now, since \(X_{1}\) and \(X_{2}\), are independent, then \(X_{1}+X_{2}\) is normal with mean \(20+30=50\) and variance \(16+9=25\)

We wish to find,

$$

\begin{align}

\text{P}\left(\text{X}_{1}+\text{X}_{2} \leq 45\right) &=\text{P}\left(\mathrm{Z}<\frac{45.5-50}{\sqrt{25}}\right) \\ &=\text{P}(\text{Z} \leq-0.9) \\

&=1-0.8159 \\ &=0.1841 \end{align}

$$

Alternatively, we can find this probability by integration,

$$

\begin{align}

\text{P}\left(\text{X}_{1}+\text{X}_{2} \leq 45\right) &=\int_{0}^{45.5} \frac{1}{\sqrt{25 \times 2 \times \pi}} \text{e}^{-\frac{1}{2}\left(\frac{\text{X}_{2}-50}{5}\right)^{2}} \\&=0.1841

\end{align}

$$

An actuary is analyzing the annual number of accidents in two neighboring cities, \(\text{P}\) and \(\text{Q}\), for its insured clients. Let \(\text{X}\) be the total annual number of accident claims from town \(\text{P}\) and let \(\text{Y}\) be the total annual number of accident claims from town \(Q\). \(X\) is normal with mean 10 and standard deviation 2. \(Y\) is also normal with mean 9 and standard deviation 2.5.

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

Calculate \(\text{P}(\text{X}+\text{Y} \geq 5)\)

Since \(\text{X}\) and \(\text{Y}\) are independent normal random variables, then \(\text{X}+\text{Y}\) is also normal with mean \(10+9=\) 19 and variance \(4+6.25=10.25\).

Now,

$$

\begin{align}

\text{P}(\text{X}+\text{Y} \geq 5) &=\text{P}\left(\text{Z}>\frac{4.5-19}{\sqrt{10.25}}\right) \\

&=\text{P}(\text{Z}>-1.874)=0.9695

\end{align}

$$

Let \(X\) and \(Y\) be the number of days of hospitalization for two individuals, \(M\) and \(N\), respectively. \(X\) and \(Y\) are normally distributed as \(X \sim N(20,9)\) and \(Y \sim N(11,7)\).

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

Find the probability that the difference between the number of days of hospitalization for \(\text{M}\) and \(\text{N}\) does not exceed \(3\).

Since \(X \sim N(20,9)\) and \(Y \sim N(11,7)\) and that \(X\) and \(Y\) are independent, then the random variable, \(\text{X}-\text{Y} \sim \text{N}(20-11,9+7) \Rightarrow \text{X}-\text{Y} \sim(9,16)\)

We wish to find,

$$\begin{align} \text{P}(\mathrm{X}-\mathrm{Y} \leq 3) &=\mathrm{P}\left(\mathrm{Z}<\frac{3.5-9}{\sqrt{16}}\right) \\

&= \mathrm{P}(\mathrm{Z}<-1.375)=1-0.9155=0.0845

\end{align}$$

*Below is the pdf for the standard normal table required for this reading*:

Please click the above icon to view the table.

**Learning Outcome**

**Topic 3. g: Multivariate Random Variables – Calculate probabilities for linear combinations of independent normal random variables**