Variance

A random variable is a variable that whose value can change under different outcomes. For example, the result of flipping a standard 6-sided die is a random variable that takes each of the values from 1 to 6 with probability 16.\frac{1}{6}.

A random variable contains a lot of information. In Expected Value, we learned that the weighted average of all possible outcomes gives one way of understanding the random variable. Recall that E[X]=μ E[X] = \mu.

The variance of a random variable measures how far the random variable deviates from its mean, by calculating the expected value of the square deviation from the mean. In mathematical symbols, we have

Var(X)=E[(X−μ)2]. Var (X) = E[ ( X - \mu)^2 ].

It can also be shown, by the linearity of expectation, that

Var(X)=E[X2]−(E[X])2. Var (X) = E[X^2] - \left( E[X] \right)^2 .

Worked Examples

1. There are 22 bags, containing balls numbered 11 through 55. From each bag, 11 ball is removed. What is the variance of the total of the two balls?

Let XX be the random variable denoting the sum of these values. Then, the probability distribution of XX is given by

x2345678910P(X=x)125225325425525425325225125\begin{array} { | l | l | l | l | l | l | l | l | l | l l | } \hline x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \\ \hline P(X=x) & \frac{1}{25} & \frac{2}{25} & \frac{3}{25} & \frac{4}{25} & \frac{5}{25} & \frac{4}{25} & \frac{3}{25} & \frac{2}{25} & \frac{1}{25} & \\ \hline \end{array}

We have previously calculated that E[X]=6. E[X] = 6. As such, we can see that

Var(X)=E[(X−μ)2]=(2−6)2×125+(3−6)2×225+(4−6)2×325+(5−6)2×425+(6−6)2×525+(7−6)2×425+(8−6)2×325+(9−6)2×225+(10−6)2×125=4. \begin{aligned} Var(X) & = E[(X - \mu)^2] \\ & = (2-6)^2 \times \frac {1}{25} + (3-6)^2 \times \frac {2}{25} + (4-6)^2 \times \frac {3}{25} \\ & + (5-6)^2 \times \frac {4}{25} + (6-6)^2 \times \frac {5}{25} + (7-6)^2 \times \frac {4}{25} \\ & + (8-6)^2 \times \frac {3}{25} + (9-6)^2 \times \frac {2}{25} + (10-6)^2 \times \frac {1}{25} \\ & = 4. \end{aligned}

 

2. 33 six-sided dice are rolled. What is the variance for the number of times a 33 is rolled? What is the variance for the total of the dice?

Let YY be the random variable representing the number of times a 33 is rolled. The below table lists the probabilities of rolling different numbers of 33s.

num 3s0123probability 12521675216152161216 \begin{array}{|c|cccc|} \hline \mbox{num 3s} & 0 & 1 & 2 & 3\\ \mbox{probability } & \frac{125}{216} & \frac{75}{216} & \frac{15}{216} & \frac{1}{216}\\ \hline \end{array}

The expected number of times a 33 is rolled is 12\frac{1}{2}, so E(Y)=12E(Y) = \frac{1}{2}. We now wish to calculate E(Y2)E(Y^2).

E(Y2)=02×125+12×75+22×15+32×1216=144216=23. E(Y^2) = \frac{0^2 \times 125 + 1^2 \times 75 + 2^2 \times 15 + 3^2 \times 1}{216} = \frac{144}{216} = \frac{2}{3}.

Therefore, Var(Y)=23−(12)2=512Var(Y) = \frac{2}{3} - \left(\frac{1}{2}\right)^2 = \frac{5}{12}.

#Combinatorics #Variance #Olympiad

Note by Calvin Lin
7 years, 2 months ago

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Comments

Why do you have to square the differences?

Nathan Ramesh - 7 years ago

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We are interested in measuring how spread out the data is. Squaring allows us to emphasize large differences while reducing the impact of small differences.

Variance, is defined as the second central moment, which is why we take it as E[(X−μ)2] E[ (X - \mu)^2] . Skew, is the third central moment, in which we take cubes instead of squares, namely E[(X−μ)3] E[ ( X - \mu)^3 ] .

Calvin Lin Staff - 7 years ago

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But those cubes won't always be positive....?

Nathan Ramesh - 7 years ago

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@Nathan Ramesh – Skew doesn't need to be a positive value. A negative skew will indicate that there (is likely to be ) a longer tail on the left.

Calvin Lin Staff - 7 years ago

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@Calvin Lin – But variance is always positive?

Nathan Ramesh - 7 years ago
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