Example 2
Suppose that 6 job assignments are randomly assigned to 5 workers. What is the probability that at least 4 of the job assignments go to the same worker?

Example 2 is left as exercise.

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Discussion of Example 1
Fix a face (say 1). Finding the probability of that at least 4 of the dice show the face 1 is a binomial problem. Then multiplying this answer by 6 will give the desired answer.

Consider obtaining a 1 as a success. Let be the number of successes when 7 dice are thrown. Then is . We have the following calculation:

Multiplying by 6 produces the desired answer.

To give some perspective to this example, for each , let be the event that at least 4 of the dice show the value of when 7 dice are rolled. The calculation above calculates the probability of the event . In this example, the event are mutually exclusive. This is the reason why we can multiply by 6 to obtained the answer in .

If we roll more dice, the event may no longer be mutually exclusive. For example, roll 8 dice and let be the event that at least 4 of the dice show the face of . Then the events are no longer mutually exclusive. To work this example, we need to use the multinomial theorem.

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Answer to Example 2

1. Compute the mean and variance of .
2. Compute the mean and variance of .
3. Compute the covariance and the correlation coefficient .

Probem 2
Let be the value of one roll of a fair die. If the value of the die is , we are given that has a binomial distribution with and (we use the notation ).

1. Compute the mean and variance of .
2. Compute the mean and variance of .
3. Compute the covariance and the correlation coefficient .

Problem 2 is left as exercise.

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Discussion of Problem 1

The joint variables and are identical to the ones in this previous post. However, we do not plan on following the approach in the previous, which is to first find the probability functions for the joint distribution and then the marginal distribution of . The calculation of covariance in Problem 1.3 can be very tedious by taking this approach.

Problem 1.1
We start with the easiest part, which is the random variable (the roll of the die). The variance is computed by .

Problem 1.2

We now compute the mean and variance of . The calculation of finding the joint distribution and then finding the marginal distribution of is tedious and has been done in this previous post. We do not take this approach here. Instead, we find the unconditional mean by weighting the conditional mean . The weights are the probabilities . The following is the idea.

We have for each . Before we do the weighting, we need to have some items about the conditional distribution . Since has a binomial distribution, we have:

For any random variable , and . The following is the second moment of , which is needed in calculating the unconditional variance .

We can now do the weighting to get the items of the variable .

Problem 1.3

The following is the definition of covariance of and :

where and .

The definition can be simplified as:

To compute , we can use the joint probability function of and to compute this expectation. But this is tedious. Anyone who wants to try can go to this previous post to obtain the joint distribution.

Note that the conditional mean is a linear function of . It is a well known result in probability and statistics that whenever a conditional mean is a linear function of , the conditional mean can be written as:

where is the mean of the respective variable, is the standard deviation of the respective variable and is the correlation coefficient. The following relates the correlation coefficient with the covariance.

Comparing and , we have and

Equating and , we have . Thus we deduce that is one-fourth of the variance of . Using , we have:

Plug in all the items of , , and into , we obtained . Both and are positive, an indication that both variables move together. When one increases, the other variable also increases. Thus makes sense based on the definition of the variables. For example, when the value of the die is large, the number of trials of is greater (hence a larger mean).

1. Discuss how the joint probability function is computed for and .
2. Compute the conditional binomial distributions where .
3. Compute the marginal probability function of and the mean and variance of .
4. Compute for all applicable and .

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Discussion of Problem 1

Problem 2 is found at the end of the post.

Problem 1.1
This is an example of a joint distribution that is constructed from taking product of conditional distributions and a marginial distribution. The marginal distribution of is a uniform distribution on the set (rolling a fiar die). Conditional of , has a binomial distribution . The following is the sample space of the joint distribution of and .

Figure 1

The joint probability function of and may be written as:

Thus the probability at each point in Figure 1 is the product of , which is , with the conditional probability , which is binomial. For example, the following diagram and equation demonstrate the calculation of

Figure 2

Problem 1.2
The following shows the calculation of the binomial distributions.

Problem 1.3
To find the marginal probability , we need to sum over all . For example, is the sum of for all . See the following diagram

Figure 3

As indicated in , each is the product of a conditional probability and . Thus the probability indicated in Figure 3 can be translated as:

We now begin the calculation.

The following is the calculation of the mean and variance of .

Problem 1.4
The conditional probability is easy to compute since it is a given that is a binomial variable conditional on a value of . Now we want to find the backward probability . Given the binomial observation is , what is the probability that the roll of the die is ? This is an application of the Bayes’ theorem. We can start by looking at Figure 3 once more.

Consider . In calculating this conditional probability, we only consider the 5 sample points encircled in Figure 3 and disregard all the other points. These 5 points become a new sample space if you will (this is the essence of conditional probability and conditional distribution). The sum of the joint probability for these 5 points is , calculated in the previous step. The conditional probability is simply the probability of one of these 5 points as a fraction of the total probability . Thus we have:

We do not have to evaluate the components that go into . As a practical matter, to find is to take each of 5 probabilities shown in and evaluate it as a fraction of the total probability . Thus we have:

Calculation of

Here’s the rest of the Bayes’ calculation:

Calculation of

Calculation of

Calculation of done earlier

Calculation of

Calculation of

Calculation of

Calculation of

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Probem 2
Let be the value of one roll of a fair die. If the value of the die is , we are given that has a binomial distribution with and (we use the notation ).

1. Discuss how the joint probability function is computed for and .
2. Compute the conditional binomial distributions where .
3. Compute the marginal probability function of and the mean and variance of .
4. Compute for all applicable and .

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Answers to Probem 2

Problem 2.3

Problem 2.4

Problem 1
Suppose there are two basketball players, each makes 50% of her free throws. In one game, player A attempted 10 free throws and player B attempted 15 free throws. Assume that the free throws of each player are independent of each other. Suppose you are told that in this game, 8 of their free throws were hits. Given this information:

1. What is the probability that player A made 4 of the hits?
2. What is the mean number of hits made by player A?
3. What is the variance of the number of hits made by player A?
4. What is the probability that player B made 5 of the hits?
5. What is the mean number of hits made by player B?
6. What is the variance of the number of hits made by player B?

Problem 2
A student took two multiple choice statistics quizzes that were independent of each other, i.e., results of one quiz did not affect the results on the other. One quiz had 8 questions and the other quiz had 10 questions. Each question had 5 choices and only one of the choices was correct. The student did not study. So she answered each question by random guessing. If the student was told that she had 5 correct answers in the two quizzes:

1. What is the probability that the student answered 3 or more questions correctly in the first quiz?
2. What is the mean number of correct answers in the first quiz?
3. What is the variance of the number of correct answers in the first quiz?
4. What is the probability that the student answered at most 3 questions correctly in the second quiz?
5. What is the mean number of correct answers in the second quiz?
6. What is the variance of the number of correct answers in the second quiz?

Refer to this post to find the background information.

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Answers

Problem 1

Problem 2

Problem 1
Suppose that the following is the probability function of the random variable .

where .

Evaluate the mean and variance of .

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Problem 2
Suppose that the following is the probability function of the random variable .

where is the probability function of the binomial distribution with  trials and probability of success , is the probability function of the binomial distribution with  trials and probability of success , and is the probability function of the binomial distribution with  trials and probability of success  .

Evaluate the mean and variance of .

Answers for Problem 2 are found at the end of the post.
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Discussion of Problem 1
The probability function is the weighted average of two binomial probability functions. The weights are and . Thus the probability distribution of is said to be the mixture of two binomial distributions with equal mixing weights.

One interpretation of the probability function is that the underlying phenomenon can be one of two phenomena. As an actuarial science example, suppose that a block of insurance policies is divided into two groups, roughly equal in size. One group is a low risk group. It has a low claim frequency (the probability of a policyholder in this group having a claim in a given year is ). The other group is a high risk group. It has a high claim frequency (the probability of a policyholder having a claim is ). Suppose you pick a policyholder at random from this block. What is the expected number of claims in a year from this randomly chosen insured? What is the variance of the number of claims?

Since the probability function is a weighted average of binomial distributions, the mean and other higher moments are the weighted average of the binomial means and higher moments. However, as you will see below, the variance of the mixture is not the weighted average of the binomial variances.

Before we work the problem, we need some preliminary facts about binomial distributions. Suppose has a binomial distribution with parameters and . This fact is denoted by the notation . Then the mean and variance of are and , respectively. Since , it follows that:

Now the calculation:

The idea for and is that the mean and the second moment of are the weighted average of the means and second moments of the binomial distributions. The following is the variance of

We use the insurance example indicated earlier to interpret the unconditional variance in . The two binomial distributions in the probability function are conditional distributions (e.g. conditional on what group of insureds the randomly chosen policyholder comes from, high risk or low risk). To put the result into perspective, note that both of these binomial distributions have the same variance, i.e., . Yet the unconditional variance is much higher than . The additional variance is the additional variance due to the uncertainty in the risk parameter of the insured (the uncertainty of what group the randomly chosen policyholder comes from).

The two binomial distributions in the probability function are conditional distributions indexed by a parameter variable that is implicit in . For example, when the randomly chosen policyholder is from the low risk group, the parameter is and the number of claims follows . When the randomly chosen policyholder is from the high risk group, the parameter is and the number of claims follows . The uncertainty in the risk parameter has the effect of increasing the unconditional variance of the mixture.

The increase in variance is a key characteristic of mixture distributions. Whenever a probability distribution is the mixture of conditional distributions, the uncertainty in the parameter variable always has the effect of increasing the unconditional variance of the mixture. In the insurance example, the uncertainty of the risk characteristics of the insureds across the entire block is reflected in the higher unconditional variance (as demonstrated in Problem 1).

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See the following blog posts for more detailed discussion of mixture distributions.

An example of a mixture

The variance of a mixture

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Answers for Problem 2

Problems
Let be a standard normal random variable.

1. Evaluate
2. Evaluate

We show that . Problem 2 is left as an exercise.

Let . The cumulative distribution function of is . We have the following.

Upon differentiation of this cdf, we have the probability density function (pdf) of .

The following is the calculation for .