Question Video: Determining the Variance for a Discrete Random Variable Mathematics

Let π denote a discrete random variable which can take the values 0, 2, and 5. Given that π has probability distribution function π(π₯) = π/(6π₯ + 6), find the variance of π.

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Video Transcript

Let π denote a discrete random variable, which can take the values zero, two, and five. Given that π has probability distribution function π of π₯ equals π over six π₯ plus six, find the variance of π.

The variance of a discrete random variable is a measure of the extent to which the values of that discrete random variable differ from their expected value. Itβs calculated using the formula variance of π is equal to the expected value of π squared minus the expected value of π squared. And we need to be clear on the difference in notation here. In the second term, weβre finding the expected value of π and we then square it, whereas in the first term weβre squaring the π-values first and then finding their expectation. We can also think of this as the expectation of the squares minus the square of the expectation.

The expected value of π and the expected value of π squared are calculated using the following formula. For the expected value of π, we multiply each π₯-value by its π of π₯ value, where π is the probability distribution function of π₯, and then we find their sum. For the expected value of π squared, we multiply each π₯ squared value by π of π₯ and then find their sum. Weβve been given the probability distribution function π of π₯, but it is in terms of an unknown value π. We know that π₯ can only take the values zero, two, and five. So these are the values in its range. The sum of all the probabilities in a probability distribution function must be equal to one.

So by finding expressions for the probabilities when π₯ equals zero, two, and five, we can then form an equation to determine the value of π. π of zero is π over six multiplied by zero plus six, which is π over six. π of two is π over six multiplied by two plus six, which is π over 18. And π of five is π over six multiplied by five plus six, which is π over 36. The sum of these three values must be equal to one, so we have the equation π over six plus π over 18 plus π over 36 is equal to one.

Writing each term on the left-hand side with a common denominator of 36 gives six π over 36 plus two π over 36 plus π over 36 is equal to one. We add the three fractions together to give nine π over 36 is equal to one. And then we can cancel a factor of nine in the numerator and denominator to give π over four equals one. To solve for π, we multiply both sides of the equation by four to give π equals four. So we found the value of π, and this allows us to determine the three probabilities exactly. π over six is four over six, or two-thirds. π over 18 is four over 18, or two-ninths. And π over 36 is four over 36, or one-ninth.

Next, weβll write the probability distribution of π₯ in a table. So we put the values π₯ can take, which are zero, two, and five, in the top row and then their probabilities, which weβve just calculated to be two-thirds, two-ninths, and one-ninth, in the second row. To find the expected value of π, we need to multiply each π₯-value by its π of π₯ value. We can add an extra row in our table for this. And we have zero multiplied by two-thirds, which is zero, two multiplied by two-ninths, which is four-ninths, and five multiplied by one-ninth, which is five-ninths.

The expected value of π is then the sum of these three values. Zero plus four-ninths plus five-ninths is nine-ninths, or simply one. To work out the expected value of π squared, we need to multiply each π₯ squared value by π of π₯. So we add a row to our table for the π₯ squared values, which are zero, four, and 25. We then add one final row to our table in which we multiply these values by the π of π₯ values. Zero multiplied by two-thirds is zero. Four multiplied by two-ninths is eight-ninths. And 25 multiplied by one-ninth is twenty-five ninths. The expected value of π squared is the sum of these three values, which is 33 over nine. And this simplifies to 11 over three.

To find the variance of π, we take this expected value of π squared, and from it we subtract the expected value of π squared. So we have 11 over three minus one squared. Thatβs 11 over three minus one, or 11 over three minus three over three, which is eight over three. So by first determining the value of this unknown constant π, which we did by recalling that the sum of all probabilities in a probability distribution function is one, we found that the variance of this discrete random variable π is eight over three.