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# Calculate probabilities and moments for linear combinations of independent random variables

Given random variables $$X_1,\ X_2,\ldots,X_p$$ and constants $$c_1, c_2,\ldots,\ c_p$$ then:

$$Y=c_1X_1+c_2X_2+\ldots+c_pX_p$$

is a linear combination of $$X_1,\ X_2,\ldots,\ X_p$$.

## Mean of a Linear Combination

If $$Y=c_1X_1+c_2X_2+\ldots+c_pX_p$$, then:

$$E\left(Y\right)=c_1E\left(X_1\right)+c_2E\left(X_2\right)+\ldots+c_pE(X_p)$$

This true because recall that, if we have  $$u\left(X,Y\right)=X+Y$$, and let’s say we want to find the mean of $$u$$. It is given by:

$$E\left[u\left(X,Y\right)\right]=E\left[X+Y\right]=E\left[X\right]+E[Y]$$

Also, note that if $$X$$ and $$Y$$ are random variables, then a linear combination of the random variables is given by:

$$Y=aX+bY$$

Where $$a$$ and $$b$$ are constants so that:

$$E(Y)=aX+bY=aE(X)+bE(Y)$$

## Variance of Linear Combinations of Random Variables

If $$X_1,\ X_2,\ \ldots,\ X_p$$ are random variables, and $$Y=c_1X_1+c_2X_2+\ldots+c_pX_p$$, then in general

$$V\left(Y\right)=c_1^2V\left(X_1\right)+c_2^2V\left(X_2\right)+\ldots+c_p^2V\left(X_p\right)+2\sum_{i<j}\sum{c_ic_jcov(X_iX_j)}$$

On the other hand if $$X_1,\ X_2,\ \ldots,\ X_p$$ are independent, then:

$$V\left(Y\right)=c_1^2V\left(X_1\right)+c_2^2V\left(X_2\right)+\ldots+c_p^2V\left(X_p\right)$$

Also, one can calculate the variance of a linear combination of r.v.s by plugging in the variances (when they are provided) of the individual r.v.s and squaring the coefficients of the random variables:

$$Var\left(aX+bY\right)=a^2Var\left(X\right)+b^2Var(Y)$$

#### Example: Mean of a Linear Combination of Random Variables

Three claims $$X$$, $$Y$$, and $$Z$$, have equal means of 70 and variances of 20. The covariance between pairs $$(X, Y)$$,  $$(X, Z)$$ and $$(Y, Z)$$ is are 1, 2, and 3 respectively. A random variable $$U$$ is defined as $$U=X+Y+Z$$.

Find $$E(U)$$ and $$Var(U)$$.

Solution

We know that:

\begin{align}E\left(U\right)&=E\left(X+Y+Z\right)\\ &=E(X)+E(Y)+E(Z)\\ & \Rightarrow E(U)=70+70+70=210 \end{align}

And for the variance:

$$Var\left(U\right)=Var\left(X\right)+Var\left(Y\right)+Var\left(Z\right)+2[Cov\left(X,Y\right)+Cov\left(X,Z\right)+Cov\left(Y,Z\right)]$$

$$=20+20+20+2\left[1+2+3\right] =72$$

## Probability of a Sum of Random Variables

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

#### Example: Probability for Linear Combinations of Independent Random Variables (Discrete Case)

A pension plan has two types of benefits, $$X$$ and $$Y$$. The benefits are independent, and the following table shows the probability mass function of $$X$$ and $$Y$$.

$$\begin{array}{c|c|c|c|c} {}&{0}&{1}&{2}&{3}\\ \hline {\text{P(x)}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\\ \hline {\text{P(y)}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\\ \end{array} Find $$P\left(X+Y=1\right)$$. Solution$$\begin{align}P\left(X+Y=1\right)&=P\left(X=1\right)\bullet P\left(Y=0\right)+P\left(X=0\right)\bullet P\left(Y=1\right)\\ &=\frac{1}{4}\bullet\frac{1}{4}+\frac{1}{4}\bullet\frac{1}{4}=\frac{1}{8}\end{align}$$#### Example: Probability for Linear Combinations of Independent Random Variables (Continous Case) A benefit plan has two types of benefits, $$X$$ and $$Y$$. The benefits are independent and have the following probability density functions:$$f(x)=\begin{cases}2e^{-2x},&2\ge 0\\ 0, &\text{elsewhere}\end{cases}f(y)=\begin{cases}2e^{-2y},&2\ge 0\\ 0, &\text{elsewhere}\end{cases}$$Find $$P(X+Y\le 4)$$. Solution Let:$$Z=X+Y$$We want to find $$P(Z\le 4)$$. Therefore, we start by finding the probability density of Z, $$f(z)$$:$$\begin{align}f\left(z\right)&=\int_{-\infty}^{\infty}{f_X\left(z-y\right)f_Y\left(y\right)dy}\\ &=\int_{0}^{z}{2e^{-2\left(z-y\right)}2e^{-2y}dy\ }\\ &=\int_{0}^{z}{4e^{-2z}dy}=\left[4ye^{-2z}\right]_0^z=4ze^{-2z}\\ \therefore f(z) &=\begin{cases}4ze^{-2z}, &\text{for }\ z\ge 0\\ 0, &\text{elsewhere}\end{cases}\end{align}$$Now,$$P\left(Z\le4\right)=\int_{0}^{4}{4ze^{-2z}dz=4\int_{0}^{4}{ze^{-2z}dz}}$$Integrating by parts, we have,$$4\int_{0}^{4}{ze^{-2z}dz}=-4\bullet\frac{1}{2}\left[ze^{-2z}\right]_0^4–4\bullet\frac{1}{2}4\int_{0}^{4}{e^{-2z}dz} = 0.99698

Learning Outcome

Topic 3.h: Multivariate Random Variables – Calculate probabilities and moments for linear combinations of independent random variables.

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