Random Variable

Concepts of Random Variable

The term "statistical experiment" is used to describe any process by which several chance observations are obtained.

All possible outcomes of an experiment comprise a set that is called the sample space. We are interested in some numerical description of the outcome.

For example, when we toss a coin $\displaystyle{3}$ times, and we are interested in the number of heads that fall, then a numerical value of $\displaystyle{0},{1},{2},{3}$ will be assigned to each sample point.

The numbers $\displaystyle{0}$, $\displaystyle{1}$, $\displaystyle{2}$, and $\displaystyle{3}$ are random quantities determined by the outcome of an experiment.

They may be thought of as the values assumed by some random variable x, which in this case represents the number of heads when a coin is tossed 3 times.

So we could write x₁ = 0, x₂ = 1, x₃ = 2 and x₄ = 3.

Definitions

1. A random variable is a variable whose value is determined by the outcome of a random experiment.
2. A discrete random variable is one whose set of assumed values is countable(arises from counting).
3. A continuous random variable is one whose set of assumed values is uncountable (arises from measurement.).

We shall use:

A capital X for the random variable and

Lower case x₁, x₂, x₃... for the values of the random variable in an experiment. These x_i then represent an event that is a subset of the sample space.

The probabilities of the events are given by: P(x₁), P(x₂), P(x₃), ...

We also use the notation $\displaystyle{P}{\left({X}\right)}$. For example, we may need to find some of the probabilities involved when we throw a die. We would write for the probability of obtaining a "5" when we roll a die as:

$\displaystyle{P}{\left({X}={5}\right)}=\frac{1}{{6}}$

 Example 1 - Discrete Random Variable
Two balls are drawn at random in succession without replacement from an urn containing $\displaystyle{4}$ red balls and $\displaystyle{6}$ black balls.
Find the probabilities of all the possible outcomes.

ANSWER:

Let X denote the number of red balls in the outcome.

Possible Outcomes	RR	RB	BR	BB
X	2	1	1	0

Here, x₁ = 2, x₂ = 1 , x₃ = 1 , x₄ = 0

Now, the probability of getting $\displaystyle{2}$ red balls when we draw out the balls one at a time is:

Probability of first ball being red $\displaystyle=\frac{4}{{10}}$

Probability of second ball being red $\displaystyle=\frac{3}{{9}}$ (because there are $\displaystyle{3}$ red balls left in the urn, out of a total of $\displaystyle{9}$ balls left.) So:

$\displaystyle{P}{\left({x}_{{1}}\right)}=\frac{4}{{10}}\times\frac{3}{{9}}=\frac{2}{{15}}$

Likewise, for the probability of red first is $\displaystyle\frac{4}{{10}}$ followed by black is $\displaystyle\frac{6}{{9}}$ (because there are $\displaystyle{6}$ black balls still in the urn and $\displaystyle{9}$ balls all together). So:

$\displaystyle{P}{\left({x}_{{2}}\right)}=\frac{4}{{10}}\times\frac{6}{{9}}=\frac{4}{{15}}$

Similarly for black then red:

$\displaystyle{P}{\left({x}_{{3}}\right)}=\frac{6}{{10}}\times\frac{4}{{9}}=\frac{4}{{15}}$

Finally, for $\displaystyle{2}$ black balls:

$\displaystyle{P}{\left({x}_{{4}}\right)}=\frac{6}{{10}}\times\frac{5}{{9}}=\frac{1}{{3}}$

As a check, if we have found all the probabilities, then they should add up to $\displaystyle{1}$.

$\displaystyle\frac{2}{{15}}+\frac{4}{{15}}+\frac{4}{{15}}+\frac{1}{{3}}=\frac{15}{{15}}={1}$

So we have found them all.

Example 2 - Continuous Random Variable

A jar of coffee is picked at random from a filling process in which an automatic machine is filling coffee jars each with $\displaystyle{1}\ \text{kg}$ of coffee. Due to some faults in the automatic process, the weight of a jar could vary from jar to jar in the range $\displaystyle{0.9}\ \text{kg}$ to $\displaystyle{1.05}\ \text{kg}$, excluding the latter.

Let X denote the weight of a jar of coffee selected. What is the range of X?

ANSWER:

Possible outcomes: 0.9 ≤ X < 1.05

That's all there is to it!