{"id":3024,"date":"2019-06-28T15:41:57","date_gmt":"2019-06-28T15:41:57","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=3024"},"modified":"2023-12-18T11:20:40","modified_gmt":"2023-12-18T11:20:40","slug":"calculate-probabilities-and-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-probabilities-and-moments-for-linear-combinations-of-independent-random-variables\/","title":{"rendered":"Calculate probabilities for linear combinations of independent normal random variables"},"content":{"rendered":"

Definition: <\/h2>\n

Let \\(X_1, X_2,\\ldots,X_n\\) be random variables and let \\(c_1, c_2,\\ldots, c_n\\) be constants.<\/p>\n

Then,<\/p>\n

$$ Y=c_1X_1+c_2X_2+\\ldots+c_nX_n $$<\/p>\n

is a linear combination<\/strong> of \\(X_1, X_2,\\ldots, X_n\\).<\/p>\n

In this reading, however, we will only base our discussion on the linear combinations of independent normal random variables.<\/p>\n

Independent Normal Random Variables<\/h2>\n

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\\).<\/p>\n

Suppose we have two random variables \\(X_1\\sim N(\\mu_1,\\sigma_1^2)\\) and \\(X_2 \\sim N(\\mu_2,\\sigma_2^2)\\), then \\(S=X_1+X_2 \\sim N\\left(\\mu_1+\\mu_2,\\sigma_1^2+\\sigma_2^2\\right)\\).<\/p>\n

We can extend the this to more than two random variables:<\/p>\n

Now, let \\(X_1, X_2,\\ldots,X_n\\) be 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, it can be proven that \\(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.<\/p>\n

Similarly, we can prove that the difference of independent Normal distributions is Normal.<\/p>\n

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

This is because:<\/p>\n

$$ E\\left[D\\right]=E\\left[X_1-X_2\\right]=E\\left[X_1\\right]-E\\left[X_2\\right]=\\mu_1-\\mu_2 $$<\/p>\n

and,<\/p>\n

$$ \\begin{align*} Var\\left(D\\right) & =Cov\\left(D,D\\right)=Var\\left(X_1\\right)-2Cov\\left(X_1,X_2\\right)+Var\\left(X_2\\right) \\\\
\n& =Var\\left(X_1\\right)-0+Var\\left(X_2\\right) \\\\
\n& =Var\\left(X_1\\right)+Var\\left(X_2\\right) \\\\ & =\\sigma_1^2+\\sigma_2^2 \\end{align*} $$<\/p>\n

Probabilities for the Sum of Independent Normal Random Variables <\/h3>\n

We can calculate the probability for linear combinations of independent normal random variables. This is illustrated in the following examples:<\/p>\n

Example 1: (SOA) <\/h4>\n

A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent. Calculate the probability that the total useful life of two randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator.<\/p>\n

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

Let \\(X_i\\) be the random variable representing the useful life of refrigerator \\(i\\), \\(i=1, 2, 3\\)<\/p>\n

\\(\\left\\{X_i\\right\\}\\) is an iid collection of r.v.\u2019s with common distribution \\(X\\sim N(\\mu=10,\\sigma^2=9)\\)<\/p>\n

We seek \\(Pr\\left(X_1+X_2 \\gt 1.9X_3\\right)=Pr\\left(X_1+X_2-1.9X_3 \\gt 0\\right)\\)<\/p>\n

Let \\(S=X_1+X_2-1.9X_3\\)<\/p>\n

Now,<\/p>\n

$$ \\begin{align*} E\\left[S\\right] & =E\\left[X_1\\right]+E\\left[X_2\\right]-E\\left[1.9X_3\\right] \\\\
\n& =E\\left[X_1\\right]+E\\left[X_2\\right]-1.9E\\left[X_3\\right] \\\\
\n& =E\\left[X\\right]+E\\left[X\\right]-1.9E\\left[X\\right] \\ \\left(\\text{Identically distributed}\\right) \\\\
\n& =10+10-1.9\\left(10\\right)=1
\n\\end{align*} $$<\/p>\n

And,<\/p>\n

$$ \\begin{align*}
\nVar\\left(S\\right) & =Var\\left(X_1\\right)+Var\\left(X_2\\right)+Var\\left(-1.9X_3\\right)\\ \\ \\ \\ (\\text{Independence}) \\\\
\n& =Var\\left(X_1\\right)+Var\\left(X_2\\right)+\\left(-1.9\\right)^2Var\\left(X_3\\right) \\\\
\n& =Var\\left(X\\right)+Var\\left(X\\right)+\\left(-1.9\\right)^2Var\\left(X\\right)\\ \\ \\ \\text{Identically Distributed} \\\\
\n& =9+9+\\left(-1.9\\right)^2\\left(9\\right)=50.49 \\end{align*} $$<\/p>\n

$$ \\therefore S\\sim N(\\mu_S=1, \\sigma_S^2=50.49) $$<\/p>\n

$$ \\begin{align*} \\Pr{\\left(S \\gt 0\\right)} &=\\Pr{\\left(\\frac{S-\\mu_S }{\\sigma_S} \\gt \\frac{0-1}{\\sqrt{50.49}}\\right)}\\ \\ \\ \\ \\ (\\text{Standardize}) \\\\
\n&=\\Pr{\\left(Z \\gt -0.1407\\ldots\\right)} \\\\
\n& =\\Pr{\\left(Z \\lt 0.1407\\ldots\\right)}\\approx0.5557 \\\\
\n\\end{align*} $$<\/p>\n

Example 2<\/h4>\n

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.<\/p>\n

Find the probability that the total annual income is exactly 6,100.<\/p>\n

Solution<\/h4>\n

Let \\(X_1, X_2,X_3\\) be the random variables for the income of the three businessmen, respectively.<\/p>\n

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\\).<\/p>\n

We wish to find,<\/p>\n

$$ P\\left(X_1+X_2+X_3=3,000\\right)=P\\left(Z=\\frac{6,100-6,000}{\\sqrt{2,500}}\\right)=P\\left(Z=2\\right) $$<\/p>\n

and from the standard normal table, <\/p>\n

$$ P\\left(Z=2\\right)=0.9772 $$<\/p>\n

Example 3<\/h4>\n

Let \\(X_1\\) and \\(X_2\\) be the random variables for the amount of sickness benefits given to two individuals, A and B, respectively. \\(X_1\\) and \\(X_2\\) are independent with the following probability density functions:<\/p>\n

$$ f\\left(x_1\\right)= \\left\\{ \\begin{matrix} \\frac{1}{32\\pi}{e^{-\\frac{1}{2}\\left(\\frac{x_1-20}{4}\\right)^2}}, & x_1 \\geq 0 \\\\ 0,& \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

and,<\/p>\n

$$ f\\left(x_2\\right)= \\left\\{ \\begin{matrix} \\frac{1}{18\\pi}{e^{-\\frac{1}{2}\\left(\\frac{x_2-20}{3}\\right)^2}}, & x_2 \\geq 0 \\\\ 0,& \\text{elsewhere} \\end{matrix} \\right. $$<\/p>\n

Find \\(P(X_1+X_2\\le 45)\\)<\/p>\n

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

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.<\/p>\n

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\\).<\/p>\n

We wish to find,<\/p>\n

$$ \\begin{align*}
\nP\\left(X_1+X_2\\le 45\\right) & =P\\left(Z<\\frac{45.5-50}{\\sqrt{25}}\\right) \\\\\n& =P\\left(Z\\le-0.9\\right) \\\\\n& =1-0.8159 \\\\\n& =0.1841\n\\end{align*} $$<\/p>\n

Alternatively, we can find this probability by integration,<\/p>\n

$$ \\begin{align*} P\\left(X_1+X_2\\le 45\\right) & =\\int_{0}^{45.5}{\\frac{1}{\\sqrt{25\\times2\\times\\pi}}e^{-\\frac{1}{2}\\left(\\frac{x_2-50}{5}\\right)^2}} \\\\
\n&=0.1841
\n\\end{align*} $$<\/p>\n

Example 4<\/h4>\n

An actuary is analyzing the annual number of accidents in two neighboring cities, \\(P\\) and \\(Q\\), for its insured clients. Let \\(X\\) be the total annual number of accident claims from town \\(P\\) and let \\(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 X and Y are independent.<\/p>\n

Calculate \\(P(X+Y\\geq 5)\\).<\/p>\n

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

Since \\(X\\) and \\(Y\\) are independent normal random variables, then \\(X+Y\\) is also normal with mean \\(10+9=19\\) and variance \\(4+6.25=10.25\\).<\/p>\n

Now,<\/p>\n

$$ \\begin{align*} P\\left(X+Y\\geq5\\right) & =P\\left(Z\\gt \\frac{4.5-19}{\\sqrt{10.25}}\\right) \\\\
\n& =P\\left(Z \\gt -1.874\\right)=0.9695
\n\\end{align*} $$<\/p>\n

Example 5<\/h4>\n

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

Find the probability that the difference between the number of days of hospitalization for \\(M\\) and \\(N\\) does not exceed 3.<\/p>\n

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

Since \\(X\\sim N(20, 9)\\) and \\(Y\\sim N(11, 7)\\) and that \\(X\\) and \\(Y\\) are independent, then the random variable,<\/p>\n

$$ X-Y\\sim N\\left(20-11, 9+7\\right)\\Rightarrow X-Y \\sim (9, 16). $$<\/p>\n

We wish to find, <\/p>\n

$$ \\begin{align*} P\\left(X-Y\\le3\\right) & =P\\left(Z<\\frac{3.5-9}{\\sqrt{16}}\\right) \\\\\n& =P(Z \\lt -1.375) \\\\\n& =1-0.9155=0.0845\n\\end{align*} $$<\/p>\n

Below is the pdf for the standard normal table required for this reading:<\/em><\/strong><\/p>\n

Standard Normal Table pdf<\/a><\/em><\/strong><\/p>\n

Please click the above icon to view the table.<\/p>\n

\n

Question<\/h2>\n

Consider two independent machines \\(A\\) and \\(B\\) producing screws. The lengths of the screws from machine A are normally distributed with a mean of 5 cm and a standard deviation of 0.1 cm. The lengths of the screws from machine B are normally distributed with a mean of 5.1 cm and a standard deviation of 0.15 cm.<\/p>\n

A quality control engineer selects one screw at random from each machine and combines them into a pair. What is the probability that the total length of the pair of screws exceeds 10.3 cm?<\/p>\n

    \n
  1. 0.1587 <\/li>\n
  2. 0.1335<\/li>\n
  3. 0.8413 <\/li>\n
  4. 0.9772 <\/li>\n
  5. 0.9956<\/li>\n<\/ol>\n

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

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

    To find the probability that the total length of the pair of screws exceeds 10.3 cm, we first need to determine the distribution of the sum of their lengths. Since the lengths are independent normal random variables, the sum will also be normally distributed. The mean of the sum will be the sum of the means, and the variance of the sum will be the sum of the variances (since standard deviation is the square root of variance and variance is additive for independent random variables).<\/p>\n

    The mean of the total length is:<\/p>\n

    $$ \\mu_{\\text{total}}=\\mu_A+\\mu_B=5+5.1=10.1\\ cm $$<\/p>\n

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

    $$ \\sigma_{\\text{total}}^2=\\sigma_A^2+\\sigma_B^2=0.01+0.0225=0.0325\\ cm^2 $$<\/p>\n

    The standard deviation of the total length is the square root of the variance:<\/p>\n

    $$ \\sigma_{\\text{total}}=\\sqrt{0.0325}=0.1803\\ cm $$<\/p>\n

    Now, we convert the question into a standard normal distribution problem by finding the Z-score for 10.3 cm:<\/p>\n

    $$ Z=\\frac{X-\\mu_{\\text{total}}}{\\sigma_{\\text{total}}}=\\frac{10.3-10.1}{0.1803}=1.11 $$<\/p>\n

    Using the standard normal distribution table, a Z-score of 1.11 corresponds to a probability of approximately 0.8665. This is the probability that the total length is less than 10.3 cm. We want the probability that the total length exceeds 10.3 cm, so we subtract this value from 1:<\/p>\n

    $$ P\\left(X \\gt 10.3\\right)=1-P\\left(X \\lt 10.3\\right)=1-0.8665=0.1335 $$<\/p>\n

    The probability that the total length of the pair of screws exceeds 10.3 cm is therefore approximately 0.1335.<\/p>\n<\/blockquote>\n

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

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

    Definition: Let \\(X_1, X_2,\\ldots,X_n\\) be random variables and let \\(c_1, c_2,\\ldots, c_n\\) be constants. Then, $$ Y=c_1X_1+c_2X_2+\\ldots+c_nX_n $$ 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…<\/p>\n","protected":false},"author":1,"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-3024","post","type-post","status-publish","format-standard","hentry","category-multivariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"\nCalculate probabilities for linear combinations of independent normal random variables - CFA, FRM, and Actuarial Exams Study Notes<\/title>\n<meta name=\"description\" content=\"The probability that the total length of a pair of screws (one from each machine) exceeds 10.3 cm is approximately 0.1335.\" \/>\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-probabilities-and-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 probabilities for linear combinations of independent normal random variables - 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