Random variables and distributions

Just a little bit of chapter 2

Random variables

On page 105, random variable \(X\) is defined to be a random process with some numerical outcome.

  • Example 0: Choose \(X\) to be the sum of the scores of game 5 between the Cleveland Cavaliers and the Golden State Warriors tonight
  • Example 1: Flip a coin and write down
    • \(X=1\) if the coin lands heads or
    • \(X=0\) if the coin lands tails
  • Example 2: Roll a standard six sided die and write down the number that comes up.
  • Example 3: Roll a 10 sided die and write down
    • \(X=1\) if the roll comes up 1, 2, or 3,
    • \(X=2\) if the roll comes up 4, 5, 6, or 7, or
    • \(X=3\) if the roll comes up 8, 9, or 10
  • Example 4: Find somebody and choose \(X\) to be their height
  • Example 5: Randomly choose a college and choose \(X\) to be the average salary of all the professors.

An important distinction:

  • Examples 0 through 3 are examples of discrete random variables because they produce only integer values.
  • Examples 4 and 5 are examples of continuous random variables because they can (in principle) produce any real number.

Distributions

For discrete random variables

On page 82, the distribution of a discrete random variable is defined to be a table of all the possible outcomes together with their probabilities. For examples 1, 2, and 3 above, the distributions are as follows:

  • Example 1: \(P(X=1) = P(X=0) = 1/2\)
  • Example 2: \(P(X=1) = P(X=2) = P(X=3) = P(X=4) = P(X=5) = P(X=6) = 1/6\)
  • Example 3:
    • \(P(X=1) = 3/10\)
    • \(P(X=2) = 4/10\)
    • \(P(X=3) = 3/10\)

Notes:

  • We’ve introduced the common notation \[P(X=x_i)=p_i\]
  • There can be any number of outcomes
  • Those outcomes need not be equally likely

Rolling two die

Suppose we roll two standard, six-sided die and add the results to get a number \(X\). Then, \(X\) is a number between 2 and 12 but they are not all equally likely. Table 2.5 on page 82 shows us the probability distribution associated with this random process. Those probabilities are

  • \(P(X=2)=1/36\)
  • \(P(X=3)=2/36\)
  • \(P(X=4)=3/36\)
  • \(P(X=5)=4/36\)
  • \(P(X=6)=5/36\)
  • \(P(X=7)=6/36\)
  • \(P(X=8)=5/36\)
  • \(P(X=9)=4/36\)
  • \(P(X=10)=3/36\)
  • \(P(X=11)=2/36\)
  • \(P(X=12)=1/36\)

The binomial distribution

A basic example

Suppose we flip a coin 5 times and count how many heads we get. This will generate a random number \(X\) between 0 and 5 but they are not all equally likely. The probabilies are:

  • \(P(X=0)=1/32\)
  • \(P(X=1)=5/32\)
  • \(P(X=2)=10/32\)
  • \(P(X=3)=10/32\)
  • \(P(X=4)=5/32\)
  • \(P(X=5)=1/32\)

Note that the probability of getting any particular sequence of 5 heads and tails is \[\frac{1}{2^5} = \frac{1}{32}.\] That explains the denominator of 32 in the list of probabilities. The numerator is the number of ways to get that value for the sum. For example, there are 10 ways to get 2 heads in 5 flips:

If we plot the possible outcomes vs their probabilites, we get something like the following:

Not too exciting just yet but we can start to see the emergence of bell curve. To really see it, we’ll need to up the flip count.

Computations with the binomial distribution

Yesterday, we discussed the binomial distribution from a primarily conceptual point of view. Today, we’ll think a bit more computationally.

Key formulae

The binomial distribution is described in section 3.4.1 of our text and everything is summarized in the box at the top of page 147, which looks something like so:

Suppose the probability of a single trial being a success is \(p\). Then the probability of observing exactly \(k\) successes in \(n\) independent trials is given by \[\begin{eqnarray} {n\choose k}p^k(1-p)^{n-k} = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} \label{binomialFormula} \end{eqnarray}\] Additionally, the mean, variance, and standard deviation of the number of observed successes are \[\begin{align} \mu &= np &\sigma^2 &= np(1-p) &\sigma &= \sqrt{np(1-p)} \label{binomialStats} \end{align}\]

The first, long and fairly complicated looking line states the formula that is actually used to compute the probability that the number of successes is equal to \(k\). We’ll use the mean and standard deviation computations in the next line later, when we relate the binomial distribution to the normal distribution.

The text also mentions the four conditions to check to verify that the random variable is binomial:

  1. The trials are independent.
  2. The number of trials, \(n\), is fixed.
  3. Each trial outcome can be classified as a or .
  4. The probability of a success, \(p\), is the same for each trial.

Examples

Example 1

I have an unfair coin that comes up heads 60% of the time. Suppose I flip that coin 10 times, count the number of heads, and call the result \(X\).

  1. What is \(P(X=3)\)?
  2. What is \(P(X<3)\)?

Solution: This is a canonical binomial distribution example. To apply the formula, we simply identify \(p\), \(n\), and \(k\).

For part (a), we have \(p=0.6\), \(n=10\), and \(k=3\). Thus,

\[P(X=7) = {10\choose 3} \times 0.6^3 \times 0.4^7 = \frac{10\times9\times8}{3\times2\times1}\times 0.0003538944 \approx 0.01061683\]

For part (b), we need to add the probabilities for \(X=0\), \(X=1\), and \(X=2\). Thus, we get \[P(X<3) = 0.4^{10} + \left({10\choose 1} \times 0.6 \times 0.4^9\right) + \left({10\choose 2} \times 0.6^2 \times 0.4^8\right) \approx 0.01229455\] Computations with R:

While it’s important to understand the forumulae behind these computations, we’ll typically do the computations on the computer. In R, we can use the choose command to compute \(n\) choose \(k\) or, better yet, we can use the dbinom command to compute the binomial distribution directly. Thus, we can compute

\[P(X=k) = \texttt{dbinom(k,n,p)}.\] Sometimes, as in part (b) of the problem, we want the probability that a random variable lies in some range - that is, \(P(a < X \leq b)\). We can compute this using the cummulative binomial distribution function pbinom. That is

\[P(X\leq k) = \sum_{i=0}^k P(X=i)= \texttt{pbinom(k,n,p)}\] and \[P(a<X\leq b) = P(X\leq b) - P(X\leq a) = \texttt{pbinom(b,n,p) - pbinom(a,n,p)}.\] As a special case, we could compute \(P(X>k)\) using \(1-P(X\leq k)\).

Example 2

Continuing with 10 flips of the unfair coin above that comes up heads 60% of the time, what is the probability that I get more than 7 heads in 10 flips?

Solution: Using the parameters above, we get

\[P(X>7) = 1-P(X\leq 7) = \texttt{1-pbinom(7,10,0.6)} \approx 0.1672898.\]

Example 3

Here’s problem 3.26 from our text:

Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.

  1. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
  2. Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
  3. What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood?
  4. What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?
  5. What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox?

Solution:

Part (a): The binomilar distribution is certainly appropriate. We have 100 trials, each with a probability of 70% independent of one another, and we want the probability of 97 successes (though, that’s an unusual way to think about contracting chicken pox). This is exactly when the binomila distribution works.

Part (b): \[P(X=97) = \texttt{dbinom(97,100,0.7)} \approx 4.117027\times10^{-12}.\] Part (c): \[P(X\neq3) = 1-\texttt{dbinom(3,100,0.7)} \approx 1-1.058683\times10^{-46},\] which is effectively 100%.

Part (d): We now reduce the number of trials to 10 \[P(X=1) = \texttt{dbinom(1,10,0.7)} \approx 0.000137781.\]

Part (e): \[1-P(X\leq3) = 1-\texttt{pbinom(3,10,0.7)} \approx 0.9894079.\]