Combinations of random variables

4. Probability Distributions: Practical 4

Combinations of random variables

In the theorical part of this course, you learned the following equations for combining random variables:

If $X = aY + bZ$ then:

$E[X] = a~E[Y] + b~E[Z]$
$Var[X] = a^2Var[Y] + b^2Var[Z] + 2ab~Cov[Y,Z]$

If $X = aY - bZ$ then:

$E[X] = a~E[Y] - b~E[Z]$
$Var[X] = a^2Var[Y] + b^2Var[Z] - 2ab~Cov[Y,Z]$

Generate two random variables $X1$ and $X2$ from two different normal distribution $N(-2, 4)$ and $N(5, 2)$ . Take $n = 10000$ for both variables. Next, create a variable $Y = -5X1 + X2$ .

Calculate the mean and variance for the combined variable $Y$ by applying the theoretical equations for combining random variables given above and the values for the population parameters of $X1$ and $X2$ .

The mean of $Y = 15$
The variance of $Y = 404$

First, you should create the random variables $X1$ and $X2$ with the function rnorm(n, mean, sd).

X1 <- rnorm(10000, -2, 4)
X2 <- rnorm(10000, 5, 2)

Now, combine these following the function as described in the question.

Y <- -5*X1 + X2

Following the equation for the combination of random variables, the mean (expected value) and variance can be calculated as follows:

$E[Y] = -5 \cdot E[X1] + E[X2]$
$Var[Y] = -5^2 \cdot Var[X1] + Var[X2] + 2 \cdot -5 \cdot Cov[X1,X2]$

You know that the respective means of $X1$ and $X2$ are $-2$ and $5$ . In addition, you know that the standard deviations are $4$ and $2$ . Recall furthermore that the variance is the squared standard deviation. The variables $X1$ and $X2$ are independent.

Therefore, you can calculate the mean and variance of $Y$ as follows:

EY <- -5*-2 + 5
VarY <- (-5)^2*4^2 + 2^2 + 2*-5*0

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