Probability Exercise 7.1.3

Question: Let \(X_1\) and \(X_2\) be independent random variables with common distribution

\[\begin{aligned} p_X = \begin{pmatrix} 0 & 1 & 2 \\ \frac{1}{8} & \frac{3}{8} & \frac{1}{2} \end{pmatrix} \end{aligned}\]

Find the distribution of the sum \(Z = X_1 + X_2\)

Solution

We can build off the method used in example 7.1 here. We can use the convolution of the above distribution to find the probability distribution for each sum value that Z can take: \(z \in \{0, 1, 2, 3, 4\}\). We’ll follow the book’s notation and let \(m(x) = p_X \forall x \in \{0,1,2\}\)

Z = 0

In the case that \(Z = 0\) both \(X_1\) and \(X_2\) are 0. We can take the below convolution:

\[\begin{aligned} P(Z = 0) = p_X(0) \cdot p_X(0) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \end{aligned}\] 
p0 <- 1/8 * 1/8

1/64 == p0
## [1] TRUE

Z = 1

This outcome could occur when either \(X_1 = 0\), and \(X_2 = 1\), or vice versa. We can take the convolution below to get the probability of the outcome \(Z=1\)

\[\begin{aligned} P(Z = 1) = m(0)m(1) + m(1)m(0) = \frac{1}{8}\frac{3}{8} + \frac{3}{8}\frac{1}{8} = \frac{6}{64} \end{aligned}\] 
p1 <- (1/8 * 3/8) + (3/8 * 1/8)

6/64 == p1
## [1] TRUE

Z = 2

This outcome could occur when either both \(X_1, X_2 = 1\), \(X_1 = 0\), and \(X_2 = 2\), or vice versa. We can take the convolution below to get the probability of the outcome \(Z=2\)

\[\begin{aligned} P(Z = 2) = m(0)m(2) + m(1)m(1) + m(2)m(0) = \frac{1}{8}\frac{1}{2} + \frac{3}{8}\frac{3}{8} + \frac{1}{2}\frac{1}{8} = \frac{17}{64} \end{aligned}\]

Checking our math with the R code below

p2 <- (1/8 * 1/2) + (3/8 * 3/8) + (1/2 * 1/8)

17/64 == p2
## [1] TRUE

Z = 3

This outcome can only occur when \(X_1 = 1\) and \(X_2 = 2\), or vice versa. Neither “input” random variable can take the value 0 and have \(Z=3\), as it’s their sum. We can take the convolution to ge \(Z\)’s distribtuion function’s value for \(Z=3\)

\[\begin{aligned} P(Z = 3) = m(1)m(2) + m(2)m(1) = \frac{3}{8}\frac{1}{2} + \frac{1}{2}\frac{3}{8} = \frac{24}{64} \end{aligned}\]

Check our math for the \(Z=3\) case:

p3 <- (3/8 * 1/2) + (1/2 * 3/8)

24/64 == p3
## [1] TRUE

Z = 4

Finally, this outcome only occurs when both \(X_1\) and \(X_2\) both are equal to 2. We can get the convolution from the probability distributions for each random variable being equal to 2 multiplied together (because \(X_1\) and \(X_2\) are independent):

\[\begin{aligned} P(Z = 4) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = \frac{16}{64} \end{aligned}\] 
p4 <- (1/2 * 1/2)

16/64 == p4
## [1] TRUE

Check

Finally, we can check our work that our probability distribution (sum of each discrete distribution value above) sums to 1:

probability_sum <- p0 + p1 + p2 + p3 + p4

1.0 == probability_sum
## [1] TRUE