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Discrete Random Variables Simplified Revision Notes

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16.1.1 Discrete Random Variables

Discrete Random Variables

A discrete value is one that can only take a certain number of fixed values (i.e., they can be counted).

A variable is random if its value depends on chance. Something can be described as random even if there is not an equal chance for each outcome.

lightbulbExample

Example: An unfair dice, which is biased, is still random.

Probability Distribution

A probability distribution is a list or a formula that describes all possible outcomes of a random variable X and the associated probabilities.

lightbulbExample

Example: The following table is the probability distribution for an unfair dice.

x123456
P(X=x)P(X = x)110\frac{1}{10}110\frac{1}{10}12\frac{1}{2}110\frac{1}{10}110\frac{1}{10}110\frac{1}{10}
  • Large X for the actual outcome!
  • Small x for possible outcomes.

Worked Example

lightbulbExample

Example: A biased dice has a probability of 34\frac{3}{4} of obtaining a 6. All other outcomes are equally likely.

State the probability distribution of XX, where XX represents the number rolled.

x123456
P(X=x)P(X = x)120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}34\frac{3}{4}

Calculations:

134=141 - \frac{3}{4} = \frac{1}{4}14÷5=520\frac{1}{4} \div 5 = \frac{5}{20}

Expectation and Variance of a D.R.V. (Discrete Random Variable)

Expectation in this context means "mean score per turn."

infoNote

Take, for example, the previous distribution:

x123456
P(X=x)P(X = x)120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}120\frac{1}{20}34\frac{3}{4}
  • E(X) means "expectation" or "mean per throw" of X.
  • E(X) = Σxp\Sigma x p In this example:
E(X)=xˉ=1×120+2×120+3×120+4×120+5×120+6×34E(X) =\bar{x}= 1 \times \frac{1}{20} + 2 \times \frac{1}{20} + 3 \times \frac{1}{20} + 4 \times \frac{1}{20} + 5 \times \frac{1}{20} + 6 \times \frac{3}{4}E(X)=214=5.25E(X) = \frac{21}{4} = 5.25
  • E(X2)E(X^2) means the expectation or mean of all the x2x^2 values.
  • E(X2)=x2ˉ=Σx2pE(X^2) = \bar{x^2} = \Sigma x^2 p In this example:
E(X2)=12×120+22×120+32×120+42×120+52×120+62×34E(X^2) = 1^2 \times \frac{1}{20} + 2^2 \times \frac{1}{20} + 3^2 \times \frac{1}{20} + 4^2 \times \frac{1}{20} + 5^2 \times \frac{1}{20} + 6^2 \times \frac{3}{4}E(X2)=29.75E(X^2) = 29.75

The variance σ2=X2ˉXˉ2\sigma^2 = \bar{X^2} - \bar{X}^2:

Var(X)=E(X2)E(X)2\text{Var}(X) = E(X^2) - E(X)^2

Thus, in this case:

Var(X)=29.755.252=1.48\text{Var}(X) = 29.75 - 5.25^2 = 1.48

This measures what we expect the variance of a set of data taken from a large number of throws to be.

Estimating Population Parameters of a Distribution

When calculating the mean and standard deviation of a set of data, in the past we have had the entire population of data for which we are calculating these statistics. In such a case, we have:

mean=μ=Σxn\text{mean}= \, \mu = \frac{\Sigma x}{n} variance=σ2=Σx2n(Σxn)2\text{variance}= \, \sigma^2 = \frac{\Sigma x^2}{n} - \left( \frac{\Sigma x}{n} \right)^2 x2ˉxˉ2\bar{x^2}-\bar{x}^2

(when we have the entire population of data)

However, we may want to estimate the mean and variance of a population using a limited sample of data. In this case, the formula for the mean stays the same, but the formula for an unbiased estimate of the variance changes. For such an estimate, we no longer use the symbol σ2\sigma^2; we use s2s^2:

s2=nn1σ2s^2 = \frac{n}{n - 1} \sigma^2

This gives an unbiased estimate of the population variance when using a sample.

Justification of Formula for s2s^2:

When calculating σ2\sigma^2, we are assuming we have n independent pieces of data. When calculating s2s^2, we use an estimate of the mean. In order to do this, the process forces our data set to have this mean. This means we have one less degree of freedom — only n1n - 1 pieces of independent data. To rectify this, we multiply σ2\sigma^2 by nn, then divide by n1n - 1 independent pieces of data we have.

lightbulbExample

Example:

σ2=Σ(xxˉ)2n\sigma^2 = \frac{\Sigma (x - \bar{x})^2}{n}

for nn independent pieces of data.

s2=Σ(xxˉ)2n1s^2 = \frac{\Sigma (x - \bar{x})^2}{n - 1}

for n1n - 1 independent pieces of data.

Thus:

s2=nn1×Σ(xxˉ)2n=nn1σ2s^2 = \frac{\cancel n}{n - 1} \times \frac{\Sigma (x - \bar{x})^2}{\cancel n} = \frac{n}{n - 1} \sigma^2
lightbulbExample

Example: In a test, class scores are as follows:

23,27,50,32,60,62,4523, 27, 50, 32, 60, 62, 45

Calculate:

a) The mean and variance of the scores for the class.

b) Estimate the mean and variance of the test scores across the country.


Solving using a calculator:

Note Summary

infoNote

Key Points:

  • When estimating the variance and mean of a population from a sample, use s2s^2 and xˉ\bar{x}.
  • When you have all the population data, use σ2\sigma^2 and μ\mu.
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