Examples of convolution (discrete case)

Consider the following problems.

Problem 1
Roll a fair die two times. What is the probability that the sum of the two rolls is 5?

Problem 2
There are two independent multiple choice quizzes where each quiz has 5 questions. Each question on the first quiz has 4 choices and each question on the second quiz has 5 choices. Suppose a student answers the questions in the quizzes by pure guessing. What is the probability that the student obtains 5 correct answers in these two quizzes?

Problem 3
For the two independent quizzes in Problem 2, find the probability function for the total number of correct answers. What is the probability that the student get 6 or more correct answers in the two quizzes?

Problem 4
There are two independent multiple choice quizzes where each quiz has 5 questions. There are 5 choices for all questions in these two quizzes. Suppose the student answers the questions in the quizzes by pure guessing. Find the probability function for the total number of correct answers.

All of the above problems are about the independent sum of discrete random variables. We demonstrate the convolution technique using Problem 2.

The Convolution Formula (Discrete Case)
Let X and Y be independent discrete random variables with probability functions p_X(x)=P(X=x) and p_Y(y)=P(Y=y), respectively. Then the following is the probability function of Z=X+Y.

\displaystyle \begin{aligned}p_Z(z)&=\sum \limits_{x+y=z} p_X(x) \ p_Y(y) \\&\text{ } \\&=\sum \limits_{x} p_X(x) \ p_Y(z-x) \\&\text{ } \\&=\sum \limits_{y} p_X(z-y) \ p_Y(y)  \end{aligned}

Note that the joint probability function of X and Y is p_{X,Y}(x,y)=p_X(x) \ p_Y(y). The convolution formula says that the probability function of the independent sum Z=X+Y is obtained by summing the joint probabiity over the line x+y=z.

Problem 2
Let X be the number of correct guesses in quiz 1 and let Y be the number of correct guesses in quiz 2. The variable X has a binomial distribution with parameters n=5 and p=\frac{1}{4}. The variable Y has a binomial distribution with parameters n=5 and p=\frac{1}{5}. Let Z=X+Y.

\displaystyle \begin{aligned}p_Z(5)&=P(X=0)P(Y=5)+P(X=1)P(Y=4)+P(X=2)P(Y=3) \\&\text{ } \\&\ \ \ +P(X=3)P(Y=2)+P(X=4)P(Y=1)+P(X=5)P(Y=0) \\&\text{ } \\&=\binom{5}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^5 \ \ \binom{5}{5} \biggl(\frac{1}{5}\biggr)^5 \biggl(\frac{4}{5}\biggr)^0 \\&\ \ \ +\binom{5}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^4 \ \ \binom{5}{4} \biggl(\frac{1}{5}\biggr)^4 \biggl(\frac{4}{5}\biggr)^1 \\&\ \ \ +\binom{5}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^3 \ \ \binom{5}{3} \biggl(\frac{1}{5}\biggr)^3 \biggl(\frac{4}{5}\biggr)^2 \\&\ \ \ +\binom{5}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^2 \ \ \binom{5}{2} \biggl(\frac{1}{5}\biggr)^2 \biggl(\frac{4}{5}\biggr)^3 \\&\ \ \ +\binom{5}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^1 \ \ \binom{5}{1} \biggl(\frac{1}{5}\biggr)^1 \biggl(\frac{4}{5}\biggr)^4 \\&\ \ \ +\binom{5}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^0 \ \ \binom{5}{0} \biggl(\frac{1}{5}\biggr)^0 \biggl(\frac{4}{5}\biggr)^5 \\&\text{ } \\&=\frac{129367}{3200000}=0.0404271875 \end{aligned}

Answers to the other problems

Problem 3
Let Z=X+Y where X and Y are as in Problem 2. The following probabilities are obtained by applying the convolution formula.

\displaystyle P(Z=0)=\frac{248832}{3200000}=0.07776

\displaystyle P(Z=1)=\frac{725760}{3200000}=0.2268

\displaystyle P(Z=2)=\frac{950400}{3200000}=0.297

\displaystyle P(Z=3)=\frac{735840}{3200000}=0.22995

\displaystyle P(Z=4)=\frac{373020}{3200000}=0.11656875

\displaystyle P(Z=5)=\frac{129367}{3200000}=0.040427

\displaystyle P(Z=6)=\frac{31085}{3200000}=0.009714

\displaystyle P(Z=7)=\frac{5110}{3200000}=0.001596875

\displaystyle P(Z=8)=\frac{550}{3200000}=0.000171875

\displaystyle P(Z=9)=\frac{35}{3200000}=0.0000109375

\displaystyle P(Z=10)=\frac{1}{3200000}

\displaystyle P(Z \ge 6)=\frac{36781}{3200000}=0.011494

Problem 4
Let Z=X+Y where both X and Y are binomial with parameter n=5 and p=\frac{1}{5}. The independent sum Z is binomial with n=10 and p=\frac{1}{5}. The following probabilities are also exercises for using the convolution formula.

\displaystyle P(Z=0)=\frac{1048576}{9765625}=0.10737

\displaystyle P(Z=1)=\frac{2621440}{9765625}=0.26844

\displaystyle P(Z=2)=\frac{2949120}{9765625}=0.30199

\displaystyle P(Z=3)=\frac{1966080}{9765625}=0.20133

\displaystyle P(Z=4)=\frac{860160}{9765625}=0.08808

\displaystyle P(Z=5)=\frac{258048}{9765625}=0.02642

\displaystyle P(Z=6)=\frac{53760}{9765625}=0.00551

\displaystyle P(Z=7)=\frac{7680}{9765625}=0.000786

\displaystyle P(Z=8)=\frac{720}{9765625}=0.0000737

\displaystyle P(Z=9)=\frac{40}{9765625}

\displaystyle P(Z=10)=\frac{1}{9765625}

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: