Apply the Central Limit Theorem to cal ...
One of the most important results in probability theory is the central limit... Read More
In the previous reading, we defined \(Y=c_1X_1+c_2X_2+\ldots+c_pX_p\) 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. We also proved that if \(X_1, X_2,\ldots,X_n\) are independent normal random variables with means \(\mu_1, \mu_2,\ldots,\mu_n\) and variances \(\sigma_1^2, \sigma_2^2,{\ldots,\sigma}_n^2,\) respectively. Then, \(X_1+ X_2+\ldots+X_n\) is normally distributed with mean \(\mu_1+ \mu_2+\ldots+\mu_n\) and variance \(\sigma_1^2+ \sigma_2^2+{\ldots+\sigma}_n^2\). This implies that the sum of independent Normal distributions is Normal. Similarly, we proved that the difference of independent Normal distributions is Normal, i.e., If \(X_1 \sim N(\mu_1,\sigma_1^2)\) and \(X_2\sim N(\mu_2,\sigma_2^2)\), then \(D=X_1-X_2 \sim N\left(\mu_1-\mu_2,\sigma_1^2+\sigma_2^2\right)\).
We can also prove that the sum of independent Poisson distributions is Poisson, i.e., if \(N_1\sim P\left(\lambda_1\right)\) and \(N_2\sim P\left(\lambda_2\right)\), then \(S=N_1+N_2\sim P\left(\lambda_1+\lambda_2\right)\).
$$ \Rightarrow E\left[S\right]=E\left[N_1+N_2\right]=E\left[N_1\right]+E\left[N_2\right] $$
Now, we may wish to calculate moments such as mean and variance for these linear combinations.
Let \(Y=c_1X_1+c_2X_2+\ldots+c_pX_p\) 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:
$$ E\left(Y\right)=c_1E\left(X_1\right)+c_2E\left(X_2\right)+\ldots+c_pE(X_p) $$
An insurance company insures farmers against losses caused to crops by hail. Let \(X\), \(Y\), and \(Z\) 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 \(X\), \(Y\), and \(Z\) are independent.
Find \(E(X+Y+Z)\).
Solution
Since \(X\), \(Y\), and \(Z\) are independent, then,
$$ \begin{align*}
\Rightarrow E\left(X+Y+Z\right) & =E\left(X\right)+E\left(Y\right)+E\left(Z\right) \\
& =20+15+25= 60 \end{align*} $$
A sickness benefit plan has two types of benefits, \(X\) and \(Y\). The benefits are independent and with the following probability density functions
$$ f\left(x\right)=\left\{ \begin{matrix} 0.2e^{-0.2x}, & x \geq 0 \\ 0, & \text{elsewhere} \end{matrix} \right. $$
$$ f\left(y\right)=\left\{ \begin{matrix} 0.1e^{-0.1y}, & y \geq 0 \\ 0, & \text{elsewhere} \end{matrix} \right. $$
Find \(E(X-Y)\)
Solution
Since \(X\) and \(Y\) are independent, then
$$ E\left(X-Y\right)=E\left(X\right)-E(Y) $$
Now,
$$ E\left(X\right)=\int_{0}^{\infty}{x\times 0.2e^{-0.2x}dx}=5 $$
and,
$$ E\left(Y\right)=\int_{0}^{\infty}{y\times 0.1e^{-0.1y}}dy=10 $$
$$ \Rightarrow E\left(X-Y\right)=5-10=-5 $$
Let \(Y=c_1X_1+c_2X_2+\ldots+c_pX_p\) 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:
$$ V\left(Y\right)=c_1^2V\left(X_1\right)+c_2^2V\left(X_2\right)+\ldots+c_p^2V\left(X_p\right) $$
The number of tornadoes experienced in Region \(X\) over the next five years follows a Poisson distribution with mean 3. For each individual tornado, the resulting loss follows an exponential distribution with mean 100. These losses remain mutually independent and are also independent of the number of tornadoes. Calculate the variance of the total loss incurred due to tornadoes in this region over the next five years.
Solution
Define \(S\) to be the total loss during the next 5 years. (Aggregate Loss Random Variable)
$$ \begin{align*} S & =X_1+X_2+\cdots+X_N \\ & =\sum_{1}^{N}X_i \ N \sim P(\lambda=3) \ \left\{X_i\right\} \text{ is iid to } X\sim Exp(\theta=100) \end{align*} $$
$$ Var\left(S\right)=E\left[Var\left(S\middle|N\right)\right]+Var(E\left[S\middle|N\right]) $$
But,
$$ \begin{align*}
Var\left(S\middle|N\right) & =Var\left(X_1+X_2+\cdots+X_N\ \right|\ N) \\
& =Var\left(X_1+X_2+\cdots+X_N\right)\ \ \ \left(\text{Independence}\right) \\
& =Var\left(X_1\right)+Var\left(X_2\right)+\cdots+Var\left(X_N\right)\ \ \ \left(\text{Independence}\right) \\
& =Var\left(X\right)+Var\left(X\right)+\cdots+Var\left(X\right)\\ & =N\times Var\left(X\right)\ (\text{Identically Distributed})
\end{align*} $$
Similarly,
$$ \begin{align*}
E\left[S\middle|N\right] & =E\left[X_1+X_2+\cdots+X_N\ \right|\ N] \\
& =E\left[X_1+X_2+\cdots+X_N\right]\ \ \ \left(\text{Independence}\right) \\
& =E\left[X_1\right]+E\left[X_2\right]+\cdots+E\left[X_N\right] \\
& =E\left[X\right]+E\left[X\right]+\cdots+E\left[X\right]=N\cdot E[X] \ \ \ \text{Identically Distributed}
\end{align*} $$
$$ \begin{align*}
\therefore Var\left(S\right) & =E\left[N\cdot Var(X)\right]+Var\left(N\cdot E\left[X\right]\right) \\ & =Var\left(X\right)\cdot E\left[N\right]+\left(E\left[X\right]\right)^2\cdot Var(N) \\
& ={100}^2\cdot 3+\left(100\right)^2\cdot 3=60,000
\end{align*} $$
An actuary wants to determine the annual number of hailstorm days in two neighboring counties, \(P\) 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:
$$ \begin{array}{c|c|c|c|c}
& 0 & 1 & 2 & 3 \\ \hline
P(x) & \dfrac{1}{3} & \dfrac{1}{6} & \dfrac{1}{3} & \dfrac{1}{6} \\ \hline
P(y) & \dfrac{1}{4} & \dfrac{1}{4} & \dfrac{1}{4} & \dfrac{1}{4}
\end{array} $$
Find \(Var(X-Y)\).
Solution
Since \(X\) and \(Y\) are independent,
$$ Var\left(X-Y\right)=Var\left(X\right)+Var(Y) $$
We know that,
$$ \begin{align*}
Var\left(X\right) & =E\left(X^2\right)-\left[E\left(X\right)\right]^2 \\
& =0^2\times \frac{1}{3}+1^2\times \frac{1}{6}+2^2\times \frac{1}{3}+3^2\times \frac{1}{6} \\ & -\left(0\times \frac{1}{3}+1\times \frac{1}{6}+2\times \frac{1}{3}+3\times \frac{1}{6}\right)^2 \\
& =\frac{11}{9}
\end{align*} $$
We know that,
$$ \begin{align*}
Var\left(Y\right) & =E\left(Y^2\right)-\left[E\left(Y\right)\right]^2 \\
& =0^2\ast\frac{1}{4}+1^2\times \frac{1}{4}+2^2\times \frac{1}{4}+3^2\times \frac{1}{4} \\ & -\left(0\times \frac{1}{4}+1\times \frac{1}{4}+2\times \frac{1}{4}+3\times \frac{1}{4}\right)^2 \\
& =\frac{5}{4}
\end{align*} $$
Therefore,
$$ Var\left(X-Y\right)=\frac{11}{9}+\frac{5}{4}=2.4722 $$
Let \(X\) and \(Y\) be the random variables for the amount of whole life insurance benefit and term insurance benefit. \(X\) and \(Y\) are independent and with the following probability density functions
$$ f\left(x\right)=\left\{ \begin{matrix} 0.02e^{-0.02x}, & x\geq 0 \\ 0, & \text{elsewhere} \end{matrix} \right. $$
$$ f\left(y\right)=\left\{ \begin{matrix} 0.01e^{-0.01y}, & y \geq 0 \\ 0, & \text{elsewhere} \end{matrix} \right. $$
Find \(Var(X+Y)\).
Solution
Since \(X\) and \(Y\) are independent, then
$$ Var(X+Y)=Var\left(X\right)+Var(Y) $$
Now,
$$ \begin{align*}
Var\left(X\right) &=E\left(X^2\right)-\left[E\left(X\right)\right]^2 \\
& =\int_{0}^{\infty}{x^2\times 0.02e^{-0.02x}dx}-\left(\int_{0}^{\infty}{x\times 0.02e^{-0.02x}dx}\right)^2 \\
&=2,500
\end{align*} $$
and,
$$ \begin{align*}
Var\left(Y\right) &=E\left(Y^2\right)-\left[E\left(Y\right)\right]^2 \\
&=\int_{0}^{\infty}{y^2\times 0.01e^{-0.01y}dy}-\left(\int_{0}^{\infty}{y\times 0.01e^{-0.01y}dy}\right)^2 \\
& =10,000
\end{align*} $$
Therefore,
$$ Var\left(X+Y\right)=2,500+10,000=12,500 $$
Question
An investment portfolio is composed of two independent assets. Asset X has an expected return of 8% with a standard deviation of 5%, and Asset Y has an expected return of 12% with a standard deviation of 10%. If the portfolio is allocated with 40% in Asset X and 60% in Asset Y, calculate the expected return and the variance of the portfolio’s return.
- Expected Return: 10.4%, Variance: 0.0080
- Expected Return: 10.4%, Variance: 0.0096
- Expected Return: 10.8%, Variance: 0.0084
- Expected Return: 10.4%, Variance: 0.004
- Expected Return: 11.2%, Variance: 0.0120
Solution
The correct answer is D.
To calculate the expected return and variance of the portfolio, we can use the following formulas for the expected return and variance of a linear combination of independent random variables:
For expected return:
$$ E\left[P\right]=w_X\cdot E\left(X\right)+w_Y\cdot E[y] $$
For variance:
$$ Var\left(P\right)=w_X^2\cdot Var\left(X\right)+w_Y^2\cdot Var\left(Y\right) $$
Where \(w_X\) and \(w_Y\) are the weights of the assets in the portfolio, \(E[X]\) and \(E[Y]\) are the expected returns of the assets, and \(Var(X)\) and \(Var(Y)\) are the variances of the assets’ returns.
Given the weights and the expected returns and variances for each asset:
$$ w_X=0.4,E\left[X\right]=0.08,\ Var\left(X\right)={0.05}^2 $$
$$
w_Y=0.6,E\left[Y\right]=0.12,\ Var\left(Y\right)={0.10}^2 $$The expected return of the portfolio is:
$$ E\left[P\right]=0.4\cdot 0.08+0.6\cdot 0.12=0.104 $$
The variance of the portfolio is:
$$ Var\left(P\right)={0.4}^2\cdot 0.0025+{0.6}^2\cdot 0.01=0.004 $$
Learning Outcome
Topic 3. h: Multivariate Random Variables – Calculate moments for linear combinations of independent random variables.