## Saturday, September 27, 2014

### Unizor - Probability - Random Variables - Problems 2

Problem 2.1.

Calculate expected value E(ξ), variance Var(ξ) and standard deviation σ(ξ) of a random variable ξ equal to a winning in a simplified game of roulette played by the following rules:

(a) You bet $1 on a number 23 in a game of roulette with a spinning wheel that includes numbers from 1 to 36, 0 and 00.

(b) If the ball stops on a spinning wheel in a cell with a number 23, you win $36.

(c) If the ball stops in any other cell (including 0 and 00), you lose your $1 bet.

Answer:

E(ξ) = −1/38 ≅ −0.026316

Var(ξ) ≅ 35.078355

σ(ξ) ≅ 5.922690

Solution

Our first step is to specify the values

x1, x2,..., xN

that our random variable ξ takes, and the corresponding probabilities

p1, p2,..., pN

of each such value.

First of all, there are only two outcomes of this game, win or lose. Correspondingly, the number N of different values our random variable can take equals to 2. So, we are dealing with two random values, 36 (win) and −1 (lose).

Considering there are 38 different outcomes for a position of ball on the spinning wheel (numbers from 1 to 36, 0 and 00) with equal chances of each, the probability of winning is 1/38, while the probability of losing is 37/38.

Here is full specification of our random variable:

P(36) = 1/38

P(−1) = 37/38

The expected value of this random variable equals to a weighted average of its values with the corresponding probabilities as weights.

E(ξ) =

36·(1/38) + (−1)·(37/38) =

= (36−37)/38 = −1/38 ≅

≅ −0.026316

The variance is a weighted average of squares of differences between the values of our random variable and its expected value with corresponding probabilities as weights.

Var(ξ) =

[36−(−1/38)]^2·(1/38) +

+ [−1−(−1/38)]^2·(37/38) ≅

≅ 35.078355

Finally, the standard deviation is a square root of the variance and equals to

σ(ξ) ≅ 5.922690

The end.

Problem 2.2.

The problem above is a particular case of a more general problem. Assume the random variable that takes only two values, X and Y with probabilities P(X)=p and P(Y)=q, where both p and q are non-negative and p+q=1.

Determine its expectation, variance and standard deviation.

Answer:

E(ξ) = X·p+Y·q

Var(ξ) = (X−Y)^2·p·q

σ(ξ) = |X−Y|·√(p·q)

Solution

There are only two outcomes of this experiment with our random variable taking values X or Y with probabilities p and q correspondingly, where p+q=1 (we will use this equality to replace 1−q with p and 1−p with q).

So, the expectation equals to

E(ξ) = X·p+Y·q

The variance, as a weighted average of squares of differences between the values of our random variable and its expectation, where weights are probabilities, is equal to

Var(ξ) =

[X−(X·p+Y·q)]^2·p +

+ [Y−(X·p+Y·q)]^2·q =

= (X−Y)^2·q^2·p + (Y−X)^2·p^2·q =

= (X−Y)^2·p·q·(p+q) =

= (X−Y)^2·p·q

Finally, the standard deviation is a square root from the variance and equals to

σ(ξ) = |X−Y|·√(p·q)

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