Mean And Variance Of Random Variables

A unique numerical value is associated with every outcome of an experiment. The value varies with every trial of the experiment. A random experiment is one whose outcome is unknown. The experiment is usually repeated under identical conditions. In an example of a coin being tossed, the outcome is a head or a tail. Every time the coin is tossed, the outcome is either of the two. The outcome is not known in prior and is unpredictable. A random variable with values being finite is a discrete random variable. A variable, which can take any value between the given limits is a continuous random variable.

How to Define Mean And Variance Of Random Variables?

The expectation of a random variable is a weighted average of all the values of a random variable. Here, the weights correspond to their respective probabilities. The formula for finding the mean of a random variable is as follows:

E (X) = μ = Σi xi pi, where i = 1, 2, …, n

E (X) = x1p1 + x2p2 + … + xnpn, where p refers to the probabilities

Variance gives the distance of a random variable from the mean. The smaller the variance, the random variable is closer to the mean.

The formula for finding the variance of a random variable is

σx2 = Var (X) = ∑i (xi − μ)2 p(xi) = E (X − μ)2 or, Var(X) = E(X2) − [E(X)]2.

E(X2) = ∑i xi2 p(xi), and [E(X)]2 = [∑i xi p(xi)]2 = μ2.

Mean And Variance Of Random Variables Examples

Examples on expectation of random variables

1] Let X represent the outcome of a roll of a fair six-sided die. More specifically, X will be the number of pips showing on the top face of the die after the toss. The possible values for X are 1, 2, 3, 4, 5, and 6, all of which are equally likely with a probability of 1/6. The expectation of X is

E[X]=116+216+316+416+516+616=3.5.{\displaystyle {E} [X]=1\cdot {\frac {1}{6}}+2\cdot {\frac {1}{6}}+3\cdot {\frac {1}{6}}+4\cdot {\frac {1}{6}}+5\cdot {\frac {1}{6}}+6\cdot {\frac {1}{6}}=3.5.}

If one rolls the die n times and computes the average (arithmetic mean) of the results, then as n grows, the average will almost surely converge to the expected value, a fact known as the strong law of large numbers.

2] The roulette game consists of a small ball and a wheel with 38 numbered pockets around the edge. As the wheel is spun, the ball bounces around randomly until it settles down in one of the pockets. Suppose random variable X represents the (monetary) outcome of a $1 bet on a single number (“straight up” bet). If the bet wins (which happens with probability 1/38 in American roulette), the payoff is $35; otherwise the player loses the bet. The expected profit from such a bet will be

E[gain from $1 bet]=$13738+$35138=$119.{\displaystyle {E} [\,{\text{gain from }}\$1{\text{ bet}}\,]=-\$1\cdot {\frac {37}{38}}+\$35\cdot {\frac {1}{38}}=-\${\frac {1}{19}}.}

That is, the bet of $1 stands to lose $119$119, so its expected value is $119.$119.-\${\frac {1}{19}} -\${\frac {1}{19}}, \text \ so \ its \ expected \ value \ is \ -\${\frac {1}{19}}.-\${\frac {1}{19}}.

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Mean And Variance Of Random Variables Problems

Example 1: If X and Y be the random variables with the distributions,

Xi 2 3 6 10
Pi 0.2 0.2 0.5 0.1
Yi 0.8 0.2 0 3 7
Pi 0.2 0.3 0.1 0.3 0.1

Then find the expected value of X and Y.

Solution:

E (x) = ∑ Xi Pi

= 2 × (0.2) + 3 × 0.2 + 6 × 0.5 + 10 × 0.1

= 0.4 + 0.6 + 3.0 + 1.0 = 5

E (Y) = ∑ Yi Pi

= (0.8) × (0.2) + (0.2) (0.3) + 0 × 0.1 + 3 × 0.3 + 7 × 0.1

= 0.16 + 0.06 + 0 + 0.9 + 0.7 = 0.38

Example 2: There are 4 numbered cards beginning with 1. Two cards need to be drawn without replacement. If x denotes the sum of the numbers on the 2 cards drawn, find the mean and variance of x.

Solution:

The possible sums are 3, 4, 5, 6, 7

x 3 4 5 6 7

P(x) 2/12 2/12 4/12 2/12 2/12

x * P(x) 3/6 4/6 5/3 6/6 7/6

x2 * P(x) 9/6 16/6 25/3 36/6 49/6

Mean = ∑ x * P(x) = 30/6 = 5

Variance = ∑ x2 * P(x) – [∑ x * P(x)]2 = [160/6] – 25 = 5/3

Example 3: Consider a meeting where a proposal is announced. Out of them, 70% of the members are in favour and 30% are against it. If a member is selected at random, taking X = 0 if he opposed, and X = 1 if he is in favour, find the mean and variance of X.

Solution:
Here, the distribution of random variable X is:
P (X = 0) = 30/100 = 0.3 and P(X = 1) = 70/100 = 0.7

X 0 1
P(X) 0.3 0.7

Mean = E (X) = 0 * 0.3 + 1 * 0.7 = 0.7

E (X2) = 02 * 0.3 + 12 * 0.7 = 0.7

Var (X) = E (X2) – [E (X)]2

= 0.7 – [0.7]2

= 0.7 – 0.49

= 0.21

Example 4: A pair of dice is thrown and let X be the random variable which represents the sum of the numbers that appear on the two dice. Find the mean of X.

Solution:
The S be the sample space when a pair of dice is thrown.
Mean And Variance Of Random Variables Examples; n(S) = 36

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The random variable X i.e. the sum of the numbers on the two dice takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.
Find Probabilities of Two Dice Thrown

X or xi 2 3 4 5 6 7 8 9 10 11 12
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Mean of the Two Dice Thrown

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