3. Overview of Discrete Random Variables

3.1 definition of a random variable

In deterministic scientific theory a variable, commonly x for location or t for time, is used to make a prediction through a mathematical model. For example, using newton’s law one can solve a differential equation which will tell you the exact location of a rocket launched into space from Cape Kennedy after, t=1 minute to t=2 minutes etc. The location will be exactly determined from the solution, based, on initial conditions of the system. There is no error in the measurements. This is classical scientific theory, and how it works. On the other hand, statistical analysis is a bit different. For example, if a pitcher in a baseball game throws a fastball every time, but it is a little different each time: sometimes the pitcher throws it as possible, other times just not quite as fast or other times puts a spin on it which causes it to sink. For the batter the pitch would not exactly known from any solution, rather it would be a bit random, or what one might call a random outcome of a statistical experiment. Each time the batter stands at the plate, the pitch coming is random. Sure it might be one of a known set – fast fastball, not quite as fast fastball, slow sinking fastball – but each event is random. Interestingly these events should be random of each other, just because the last pitch was a fast fastball doesn’t mean the next one won’t be.

In statistical analysis we define a random variable, RV, to be a mathematical formalization of an outcome of a statistical experiment which depends on random events. The value of x is commonly used, and if the corresponding probability density is defined on real numbers then

3.2 discrete probability distributions & examples

It is common for a random variable x, which has n possible outcomes

that have the corresponding probabilities

e.g .

to be presented in a chart as

[latex]\begin{matrix}x_1&p_1\\x_2&p_2\\:&:\\\end{matrix}[/latex]

If the outcomes of an experiment, which is the purchasing of a lottery ticket, which has only three outcomes:

or

with the corresponding probabilities chart as

[latex]\begin{matrix}-10&0.95\\90&0.04999\\990&0.00001\\\end{matrix}[/latex]

Example 3.2.1

Find the expected value of the

[latex]\begin{matrix}-10&0.95\\90&0.04999\\990&0.00001\\\end{matrix}[/latex]

and interpret what this result can tell about the experiment.

The above formula

Rather than discuss many repetitive examples of this, let us now consider examples of one of the most useful discrete probability distributions, the binomial.

where the random variable considered in the experiment has only two possible outcome, a success with associated probability = p, or a failure with associated probability = 1-p. Moreover, the experiment under consideration is repeated n times, with the trails being truly independent so that the result of the last trial has no effect on the result nor probabilities for the current trial.

[latex]VAR=\sum_{i=1}^{n}{\left(x_i-\mu\right)^2\bullet p_i}[/latex]

where μ is the numerical result obtained as the expected value.

It is worthy to note here that the factorials term out front, often called nCx or “n choose x,” is often done in a separate computation, for example using an online calculator, so it is more common to see the binomial written as

[latex]nCx\bullet\ p^x\bullet\left(1-p\right)^{n-x}[/latex]

Example 3.2.1

Use a binomial probability distribution to find the probability of getting 7 answers correct from 10 total questions on a multiple choice test where each question has four choices, e.g. the probability of a correct guess is 1 out of 4,

Now, the solution here would be obtained from the binomial

[latex]nCx\bullet\ p^x\bullet\left(1-p\right)^{n-x}[/latex]

plugging in p as 0.25 and n = 10 which yields our density function

[latex]_{10}C_x\bullet{0.25}^x\bullet\left(0.75\right)^{10-x}.[/latex]

The desired solution is obtained by plugging in x as 3 which yields, noting that ₁₀C₃ is found to be 120 from the calculator, the solution

[latex]_{10}C_3\bullet{0.25}^7\bullet\left(0.75\right)^3\approx0.003[/latex]

or 0.3%.

It is worthy to note that the solution of the prior example, 0.3%, tells us the probability to get exactly 7 right from 10 guess. If passing the test is defined as getting seven or more right, our solution is not the probability of passing. Rather to find such a value we would need to first use the formula again to find the probability of getting 8 right, p₈, and then find the probability of getting 9 right, p₉, and then probability of getting them all right, p₁₀, hence

[latex]P\left(win\right)=p_7 p_8 p_9 p_{10}[/latex]

It is worthy to note that in practice one would not prefer to perform this calculation, and since it is possible to approximate our binomial with a regular normal, having mean =μ, and variance= σ², the same solution there could be computed as [latex]P(7 < x < 10)[/latex] using the normal density.

[latex]P\left( X \right) = \frac{{\lambda^{x} {e^{ - \lambda }}}}{{x!}}[/latex]

Once this function is defined for the probability experiment under consideration, then the remaining computations are logically the same as those outlined in the prior examples using the binomial. For example, if it is historically known that house on the intercostal river floods once every 50 years on average, then lambda would be 1 and we could set up our function as

[latex]P\left( X \right) = \frac{{1^{x} {e^{ - 1 }}}}{{x!}}[/latex]

Then, we could use it to compute the probability of no floods in the next 50 years, i.e. put x as zero, to be

[latex]P\left( {X = 0} \right) = \frac{{{1^0}{e^{ - 1}}}}{{0!}} \approx 37\% [/latex]

Likewise if we knew at our local airport historically a flight, which flew daily Monday through Friday, arrived more than 15 minutes late 2 times out of the week, then lambda would be 2 and  we could set up our function as

[latex]P\left( X \right) = \frac{{2^{x} {e^{ - 2 }}}}{{x!}}[/latex]

which we could use compute various probabilities. In these applications, along with many other probability applications, it is very important to understand the implications of the phrase “are independently of the time since,” which is basically saying that each day is a new day. A good example is the river example, let us say that the river flooded last year and ponder the question if that has any affect on the likelihood of it flooding this year.  While our common sense may make us think, well if it flooded last year then it most likely will not flood this year, this does not agree with what probability tells us. Using the Poisson probability density from our river example, and plugging in x as two, i.e. finding the probability of two floods in fifty years, we find the probability to be 

[latex]P\left( {X =2} \right) = \frac{{{1^2}{e^{ - 1}}}}{{2!}} \approx 18\%[/latex]

This tells us that there is an eighteen percent chance that this river will flood again in the next forty-nine years, but it does not tell us anything about when this will occur. It is equal likely to occur this year as it is occur next year or the following year, and so forth. While this concept may not agree with our common sense, it is how probability works when we have the assumption of independence. Of course, not all probability problems have the assumption of independence and there is a procedure called conditional probability that addresses problems where the likelihood of the next event occurring does depend on results of prior outcomes.