{"id":28526,"date":"2022-09-01T09:21:59","date_gmt":"2022-09-01T09:21:59","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=28526"},"modified":"2023-12-18T11:41:39","modified_gmt":"2023-12-18T11:41:39","slug":"calculate-moments-for-linear-combinations-of-independent-random-variables","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/calculate-moments-for-linear-combinations-of-independent-random-variables\/","title":{"rendered":"Calculate moments for linear combinations of independent random variables"},"content":{"rendered":"

In the previous reading, we defined \\(Y=c_1X_1+c_2X_2+\\ldots+c_pX_p\\) to be a linear combination<\/strong> 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)\\).<\/p>\n

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)\\).<\/p>\n

$$ \\Rightarrow E\\left[S\\right]=E\\left[N_1+N_2\\right]=E\\left[N_1\\right]+E\\left[N_2\\right] $$<\/p>\n

Now, we may wish to calculate moments such as mean and variance for these linear combinations. <\/p>\n

Mean of a Linear Combination of Independent Random Variables<\/h2>\n

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:<\/p>\n

$$ E\\left(Y\\right)=c_1E\\left(X_1\\right)+c_2E\\left(X_2\\right)+\\ldots+c_pE(X_p) $$<\/p>\n

Example 1: Mean of a Linear Combination of Independent Random Variables<\/h4>\n

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.<\/p>\n

Find \\(E(X+Y+Z)\\).<\/p>\n

Solution <\/strong><\/p>\n

Since \\(X\\), \\(Y\\), and \\(Z\\) are independent, then,<\/p>\n

$$ \\begin{align*}
\n \\Rightarrow E\\left(X+Y+Z\\right) & =E\\left(X\\right)+E\\left(Y\\right)+E\\left(Z\\right) \\\\
\n& =20+15+25= 60 \\end{align*} $$<\/p>\n

Example 2: Mean of a Linear Combination of Independent Random Variables<\/h4>\n

A sickness benefit plan has two types of benefits, \\(X\\) and \\(Y\\). The benefits are independent and with the following probability density functions<\/p>\n

$$ f\\left(x\\right)=\\left\\{ \\begin{matrix} 0.2e^{-0.2x}, & x \\geq 0 \\\\ 0, & \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

$$ f\\left(y\\right)=\\left\\{ \\begin{matrix} 0.1e^{-0.1y}, & y \\geq 0 \\\\ 0, & \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

Find \\(E(X-Y)\\)<\/p>\n

Solution<\/strong><\/p>\n

Since \\(X\\) and \\(Y\\) are independent, then<\/p>\n

$$ E\\left(X-Y\\right)=E\\left(X\\right)-E(Y) $$<\/p>\n

Now,<\/p>\n

$$ E\\left(X\\right)=\\int_{0}^{\\infty}{x\\times 0.2e^{-0.2x}dx}=5 $$<\/p>\n

and,<\/p>\n

$$ E\\left(Y\\right)=\\int_{0}^{\\infty}{y\\times 0.1e^{-0.1y}}dy=10 $$<\/p>\n

$$ \\Rightarrow E\\left(X-Y\\right)=5-10=-5 $$<\/p>\n

Variance of a Linear Combination of Independent Random Variables<\/h2>\n

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:<\/p>\n

$$ V\\left(Y\\right)=c_1^2V\\left(X_1\\right)+c_2^2V\\left(X_2\\right)+\\ldots+c_p^2V\\left(X_p\\right) $$<\/p>\n

Typical SOA Exam Question<\/h4>\n

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.<\/p>\n

Solution<\/strong><\/p>\n

Define \\(S\\) to be the total loss during the next 5 years. (Aggregate Loss Random Variable)<\/p>\n

$$ \\begin{align*} S & =X_1+X_2+\\cdots+X_N \\\\ & =\\sum_{1}^{N}X_i \\ N \\sim P(\\lambda=3) \\ \\left\\{X_i\\right\\} \\text{ is iid to } X\\sim Exp(\\theta=100) \\end{align*} $$<\/p>\n

$$ Var\\left(S\\right)=E\\left[Var\\left(S\\middle|N\\right)\\right]+Var(E\\left[S\\middle|N\\right]) $$<\/p>\n

But,<\/p>\n

$$ \\begin{align*}
\n Var\\left(S\\middle|N\\right) & =Var\\left(X_1+X_2+\\cdots+X_N\\ \\right|\\ N) \\\\
\n& =Var\\left(X_1+X_2+\\cdots+X_N\\right)\\ \\ \\ \\left(\\text{Independence}\\right) \\\\
\n& =Var\\left(X_1\\right)+Var\\left(X_2\\right)+\\cdots+Var\\left(X_N\\right)\\ \\ \\ \\left(\\text{Independence}\\right) \\\\
\n& =Var\\left(X\\right)+Var\\left(X\\right)+\\cdots+Var\\left(X\\right)\\\\ & =N\\times Var\\left(X\\right)\\ (\\text{Identically Distributed})
\n\\end{align*} $$<\/p>\n

Similarly,<\/p>\n

$$ \\begin{align*}
\n E\\left[S\\middle|N\\right] & =E\\left[X_1+X_2+\\cdots+X_N\\ \\right|\\ N] \\\\
\n& =E\\left[X_1+X_2+\\cdots+X_N\\right]\\ \\ \\ \\left(\\text{Independence}\\right) \\\\
\n& =E\\left[X_1\\right]+E\\left[X_2\\right]+\\cdots+E\\left[X_N\\right] \\\\
\n& =E\\left[X\\right]+E\\left[X\\right]+\\cdots+E\\left[X\\right]=N\\cdot E[X] \\ \\ \\ \\text{Identically Distributed}
\n\\end{align*} $$<\/p>\n

$$ \\begin{align*}
\n\\therefore Var\\left(S\\right) & =E\\left[N\\cdot Var(X)\\right]+Var\\left(N\\cdot E\\left[X\\right]\\right) \\\\ & =Var\\left(X\\right)\\cdot E\\left[N\\right]+\\left(E\\left[X\\right]\\right)^2\\cdot Var(N) \\\\
\n& ={100}^2\\cdot 3+\\left(100\\right)^2\\cdot 3=60,000
\n\\end{align*} $$<\/p>\n

Example 3: Variance of a Linear Combination of Independent Random Variables<\/h4>\n

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:<\/p>\n

$$ \\begin{array}{c|c|c|c|c}
\n&\t0 &\t1 &\t2 &\t3 \\\\ \\hline
\nP(x) &\t\\dfrac{1}{3} &\t\\dfrac{1}{6} &\t\\dfrac{1}{3} &\t\\dfrac{1}{6} \\\\ \\hline
\nP(y) &\t\\dfrac{1}{4} &\t\\dfrac{1}{4} &\t\\dfrac{1}{4} &\t\\dfrac{1}{4}
\n\\end{array} $$<\/p>\n

Find \\(Var(X-Y)\\).<\/p>\n

Solution<\/strong><\/p>\n

Since \\(X\\) and \\(Y\\) are independent, <\/p>\n

$$ Var\\left(X-Y\\right)=Var\\left(X\\right)+Var(Y) $$<\/p>\n

We know that,<\/p>\n

$$ \\begin{align*}
\nVar\\left(X\\right) & =E\\left(X^2\\right)-\\left[E\\left(X\\right)\\right]^2 \\\\
\n& =0^2\\times \\frac{1}{3}+1^2\\times \\frac{1}{6}+2^2\\times \\frac{1}{3}+3^2\\times \\frac{1}{6} \\\\ & -\\left(0\\times \\frac{1}{3}+1\\times \\frac{1}{6}+2\\times \\frac{1}{3}+3\\times \\frac{1}{6}\\right)^2 \\\\
\n& =\\frac{11}{9}
\n\\end{align*} $$<\/p>\n

We know that,<\/p>\n

$$ \\begin{align*}
\nVar\\left(Y\\right) & =E\\left(Y^2\\right)-\\left[E\\left(Y\\right)\\right]^2 \\\\
\n& =0^2\\ast\\frac{1}{4}+1^2\\times \\frac{1}{4}+2^2\\times \\frac{1}{4}+3^2\\times \\frac{1}{4} \\\\ & -\\left(0\\times \\frac{1}{4}+1\\times \\frac{1}{4}+2\\times \\frac{1}{4}+3\\times \\frac{1}{4}\\right)^2 \\\\
\n& =\\frac{5}{4}
\n\\end{align*} $$<\/p>\n

Therefore,<\/p>\n

$$ Var\\left(X-Y\\right)=\\frac{11}{9}+\\frac{5}{4}=2.4722 $$<\/p>\n

Example 4: Variance of a Linear Combination of Independent Random Variables<\/h4>\n

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<\/p>\n

$$ f\\left(x\\right)=\\left\\{ \\begin{matrix} 0.02e^{-0.02x}, & x\\geq 0 \\\\ 0, & \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

$$ f\\left(y\\right)=\\left\\{ \\begin{matrix} 0.01e^{-0.01y}, & y \\geq 0 \\\\ 0, & \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

Find \\(Var(X+Y)\\).<\/p>\n

Solution<\/strong><\/p>\n

Since \\(X\\) and \\(Y\\) are independent, then<\/p>\n

$$ Var(X+Y)=Var\\left(X\\right)+Var(Y) $$<\/p>\n

Now,<\/p>\n

$$ \\begin{align*}
\nVar\\left(X\\right) &=E\\left(X^2\\right)-\\left[E\\left(X\\right)\\right]^2 \\\\
\n& =\\int_{0}^{\\infty}{x^2\\times 0.02e^{-0.02x}dx}-\\left(\\int_{0}^{\\infty}{x\\times 0.02e^{-0.02x}dx}\\right)^2 \\\\
\n&=2,500
\n\\end{align*} $$<\/p>\n

and,<\/p>\n

$$ \\begin{align*}
\nVar\\left(Y\\right) &=E\\left(Y^2\\right)-\\left[E\\left(Y\\right)\\right]^2 \\\\
\n&=\\int_{0}^{\\infty}{y^2\\times 0.01e^{-0.01y}dy}-\\left(\\int_{0}^{\\infty}{y\\times 0.01e^{-0.01y}dy}\\right)^2 \\\\
\n& =10,000
\n\\end{align*} $$<\/p>\n

Therefore,<\/p>\n

$$ Var\\left(X+Y\\right)=2,500+10,000=12,500 $$<\/p>\n

\n

Question<\/h2>\n

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.<\/p>\n

    \n
  1. Expected Return: 10.4%, Variance: 0.0080 <\/li>\n
  2. Expected Return: 10.4%, Variance: 0.0096 <\/li>\n
  3. Expected Return: 10.8%, Variance: 0.0084 <\/li>\n
  4. Expected Return: 10.4%, Variance: 0.004 <\/li>\n
  5. Expected Return: 11.2%, Variance: 0.0120<\/li>\n<\/ol>\n

    Solution<\/strong><\/p>\n

    The correct answer is D.<\/strong><\/p>\n

    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:<\/p>\n

    For expected return:<\/p>\n

    $$ E\\left[P\\right]=w_X\\cdot E\\left(X\\right)+w_Y\\cdot E[y] $$<\/p>\n

    For variance:<\/p>\n

    $$ Var\\left(P\\right)=w_X^2\\cdot Var\\left(X\\right)+w_Y^2\\cdot Var\\left(Y\\right) $$<\/p>\n

    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.<\/p>\n

    Given the weights and the expected returns and variances for each asset:<\/p>\n

    $$ w_X=0.4,E\\left[X\\right]=0.08,\\ Var\\left(X\\right)={0.05}^2 $$<\/p>\n

    $$
    \nw_Y=0.6,E\\left[Y\\right]=0.12,\\ Var\\left(Y\\right)={0.10}^2 $$<\/p>\n

    The expected return of the portfolio is: <\/p>\n

    $$ E\\left[P\\right]=0.4\\cdot 0.08+0.6\\cdot 0.12=0.104 $$<\/p>\n

    The variance of the portfolio is:<\/p>\n

    $$ Var\\left(P\\right)={0.4}^2\\cdot 0.0025+{0.6}^2\\cdot 0.01=0.004 $$<\/p>\n<\/blockquote>\n

    Learning Outcome<\/strong><\/em><\/p>\n

    Topic 3. h: Multivariate Random Variables – Calculate moments for linear combinations of independent random variables.<\/strong><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

    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…<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[99],"tags":[],"class_list":["post-28526","post","type-post","status-publish","format-standard","hentry","category-multivariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"\nCalculate moments for linear combinations of independent random variables - CFA, FRM, and Actuarial Exams Study Notes<\/title>\n<meta name=\"description\" content=\"Explore the concept of linear combinations of independent random variables, including the mean and variance calculations.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/calculate-moments-for-linear-combinations-of-independent-random-variables\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Calculate moments for linear combinations of independent random variables - 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