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Calculate probabilities and moments for linear combinations of independent random variables

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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