6.2: Applications of Discrete Random Variables
- Page ID
- 105838
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Section 1: Expected Value and Variance
The mean of a discrete random variable \(X\), denoted by \(E[X]\) or \(\mu_X\), is
\(\sum x_i P(X=x_i)\)
Another name for the mean is the expected value.
The variance of a discrete random variable \(X\), denoted by \(VAR[X]\) or \(\sigma_{X}^2\), is
\(\sum (x_i-\mu_X)^2\cdot P(X=x_i)\)
and the standard deviation of a discrete random variable \(X\), denoted by \(SD[X]\) or \(\sigma_X\), is
\(\sqrt{VAR[X]}\)
Let \(Y\) be the number of cars owned by a randomly selected student with the following probability distribution table for \(Y\):
|
\(Y\) |
\(y_i\) |
\(P(Y=y_i)\) |
|---|---|---|
|
0 |
0.24 |
|
|
1 |
0.64 |
|
|
2 |
0.08 |
|
|
3 |
0.00 |
|
|
4 |
0.04 |
|
|
Total: |
1 |
Solution
To compute the expected value of \(Y\) we can use the probability distribution table in the following way:
- We add another column on the right in which for each outcome we write the product of the outcome by its probability.
- Finally, we find the total 0.96.
|
\(Y\) |
\(y_i\) |
\(P(Y=y_i)\) |
\(y_i\cdot P(Y=y_i)\) |
|---|---|---|---|
|
0 |
0.24 |
\(0\cdot0.24=0\) |
|
|
1 |
0.64 |
\(1\cdot0.64=0.64\) |
|
|
2 |
0.08 |
\(2\cdot0.08=0.16\) |
|
|
3 |
0.00 |
\(3\cdot0.00=0\) |
|
|
4 |
0.04 |
\(4\cdot0.04=0.16\) |
|
|
Total: |
1 |
\(0.96\) |
The total is labeled \(\mu_Y\), the mean of \(Y\), or \(E[Y]\), the expected value of \(Y\).
To compute the variance of \(Y\) we modify the probability distribution table in the following way.
- First, we compute the expected value as previously discussed.
- Next, we add another column on the right in which for each outcome we write the squared difference between each outcome and the expected value.
- Then, we add another column on the right in which for each outcome we write the product of the squared difference by the probability of the outcome.
- Finally, we find the total 0.6784 and the square root of the total \(\sqrt{0.6784}=0.82\).
|
\(Y\) |
\(y_i\) |
\(P(Y=y_i)\) |
\((y_i-E[Y])^2\) |
\((y_i-E[Y])^2\cdot P(Y=y_i)\) |
|---|---|---|---|---|
|
0 |
0.24 |
\((0-0.96)^2=0.9216\) |
\(0.9216\cdot0.24=0.2212\) |
|
|
1 |
0.64 |
\((1-0.96)^2=0.0016\) |
\(0.0016\cdot0.64=0.0010\) |
|
|
2 |
0.08 |
\((2-0.96)^2=1.0816\) |
\(1.0816\cdot0.08=0.0865\) |
|
|
3 |
0.00 |
\((3-0.96)^2=4.1616\) |
\(4.1616\cdot0.00=0.0000\) |
|
|
4 |
0.04 |
\((4-0.96)^2=9.2416\) |
\(9.2416\cdot0.04=0.3697\) |
|
|
\(E[Y]=0.96\) |
1 |
\(VAR[Y]=0.6784\) |
||
|
\(SD[Y]=\sqrt{0.6784}=0.82\) |
The total is labeled \(\sigma_Y^2\), the variance of \(Y\), and its square root is labeled \(\sigma_Y\), the standard deviation of \(Y\).
In summary, for a discrete random variable \(X\) we can compute the mean and the standard deviation. The process can be summarized with the following formulas:
|
Expected Value, \(E[X]=\mu_X\) |
Standard Deviation, \(SD[X]=\sigma_X\) |
|---|---|
|
\(\mu_X=\sum x_i P(X=x_i)\) |
\(\sigma_X=\sqrt{\sum (x_i-\mu_X)^2\cdot P(X=x_i)}\) |
|
We interpret the mean of \(X\) as the average value of the variable in a long run. In other words, if we perform the experiment many times, the average value of many outcomes can be predicted by the mean of \(X\). So the mean of \(X\) sets up the “expectations” from the experiment before the experiment is actually performed. |
We interpret the standard deviation of \(X\) as the expected deviation of the outcomes from the expected value in a long run. |
However, the standard deviation is better interpreted when combined with the mean of \(X\) using the three standard deviations rule in the following way:
|
Range |
Interpretation |
|---|---|
|
within \(2\sigma_x\) from \(\mu_X\) |
“expected” |
|
more than \(2\sigma_x\) from \(\mu_X\) |
“unusual” |
|
more than \(3\sigma_x\) from \(\mu_X\) |
“abnormal” |
For a discrete random variable \(X\) with \(\mu_X=10\) and \(\sigma_X=2\) then by the three standard deviations rule, we may conclude that:
|
Range |
Numerical Range |
Interpretation |
|---|---|---|
|
within \(2\sigma_x\) from \(\mu_X\) |
between 6 and 14 |
“expected” |
|
more than \(2\sigma_x\) from \(\mu_X\) |
less than 6 or more than 14 |
“unusual” |
|
more than \(3\sigma_x\) from \(\mu_X\) |
less than 4 or more than 16 |
“abnormal” |
For \(Y\), the number of cars owned by a randomly selected student, with \(\mu_Y=0.96\) and \(\sigma_Y=0.82\), we may conclude that:
|
Range |
Numerical Range |
Interpretation |
|---|---|---|
|
within \(2\sigma_x\) from \(\mu_X\) |
between -0.68 and 2.60 |
“expected” |
|
more than \(2\sigma_x\) from \(\mu_X\) |
less than -0.68 or more than 2.60 |
“unusual” |
|
more than \(3\sigma_x\) from \(\mu_X\) |
less than -1.50 or more than 3.42 |
“abnormal” |
Section 2: Applications of Discrete Random Variables
Gambling
Let’s play an American roulette! In American roulette there are 18 red slots, 18 black slots, and 2 green slots.
So, the probability of the ball landing on red is 18/38 and the probability of the ball landing on green or black is 20/38. The rules of the game are such that if you place a $1 bet on red, you will win a $1 if the result of the spin is red and lose $1 if the result of the spin is not red. Placing a bet can be thought of as an experiment whose outcome depends on a chance, therefore we can introduce a random variable \(X\) representing the payoff of such a bet. Then the following probability distribution table can be constructed for \(X\) by listing the possible values of \(X\), for winning and losing scenarios, along with their probabilities.
|
\(X\) |
outcome |
\(x_i\) |
\(P(X=x_i)\) |
|---|---|---|---|
|
Red |
+1 |
18/38 |
|
|
Black or Green |
-1 |
20/38 |
|
|
Total: |
1 |
We can now compute the expected value by adding another column on the right and computing the product of each possible value by its probability. The sum of the column is \(-\frac{2}{38}=-0.05\) and this is the value that we call the expected value.
|
\(X\) |
outcome |
\(x_i\) |
\(P(X=x_i)\) |
\(x_i\cdot P(X=x_i)\) |
|---|---|---|---|---|
|
Red |
+1 |
18/38 |
18/38 |
|
|
Black or Green |
-1 |
20/38 |
-20/38 |
|
|
Total: |
1 |
-2/38 |
So what does this number mean in the context of this problem? We interpret the mean as the average winnings of placing a $1 bet on red! In other words, on average you would lose 5 cents per game. Of course, every game you either win or lose a dollar but on average it is always a loss! Every game is set up in such a way that the expected payoff of a bet is negative! Do not gamble!
Insurance
Another application of a probability is in the insurance industry. Let’s say Bobert just purchased a new car and wants to purchase an additional insurance policy against totaling the car. So Bobert goes to insurance agent’s office and says, “I want to purchase a policy against totaling my $20000 car”. What happens next? The agent would collect the data about Bobert such as age, driving experience etc. and send it to an actuary. Actuary is a person whose entire job is to evaluate the probabilities of different events. In this case, the actuary will look at the data on how many people just like Bobert had totaled their cars annually in the past and using the data they can estimate the probability of Bobert totaling his car in the next year. So, let us say the probability of him totaling his car during in the next year is estimated to be 3.5%. Using the complimentary rule, we can compute the probability of him not totaling the car which is 96.5%. Let’s say that Bobert ended up purchasing the policy for a lump sum of $1000 called premium, and in exchange he received an assurance that in case of him totaling the car he will get paid $20000.
The purchase of an insurance can be thought of as an experiment whose outcome depends on a chance, therefore we can introduce a random variable \(X\) representing the profit of the insurance company from selling the insurance to Bobert. Then the following probability distribution table can be constructed for \(X\) by listing the possible values of \(X\), for not totaling and totaling scenarios, along with their probabilities.
|
\(X\) |
outcome |
\(x_i\) |
\(P(X=x_i)\) |
|---|---|---|---|
|
Not Totaling |
+1000 |
0.965 |
|
|
Totaling |
-20000+1000 |
0.035 |
|
|
Total: |
1 |
We can now compute the expected value by adding another column on the right and computing the product of each possible value by its probability. The sum of the column is 300 and this is the value that we call the expected value.
|
\(X\) |
outcome |
\(x_i\) |
\(P(X=x_i)\) |
\(x_i\cdot P(X=x_i)\) |
|---|---|---|---|---|
|
Not Totaling |
+1000 |
0.965 |
965 |
|
|
Totaling |
-20000+1000 |
0.035 |
-665 |
|
|
Total: |
1 |
300 |
So what does this number mean in the context of this problem? We interpret the mean as the average profit of the insurance company from insuring drivers like Bobert. From insuring 1000 people like Bobert they expect to make 300,000 dollars.
In summary, buying an insurance policy is equivalent to making a bet with the insurance company that a certain event will happen and insurance company would not sell you a policy unless they had a positive expected profit. Insurance business is a very profitable business.