Problem # 1: An urn contains 5 white balls, 3 black balls and 2 red balls. Exper

Question

Problem # 1: An urn contains 5 white balls, 3 black balls and 2 red balls. Experiment # 1 : A single ball is drawn at random from the urn. Let X be the random variable that is 1 if the ball is black, and is 0 otherwise. Let Y be the random variable that is 1 if the ball is white or black, and is 0 otherwise. (a) Find the variance of X. (b) Find the variance of Y (c) Find the covariance of X and Y. (Note that the covariance of X and Y is defined as Cov(X,Y) E[ (X EX)) (Y- EM)]) (d) Interpret your answer to Part (c). (Do you expect the covariance to be 0 or non-zero? Positive or negative? Why?) Experiment # 2: We keep drawing a single ball at random from the urn, with replacement, until we draw a red ball. (We stop when we draw the red ball.) Let Z be the number of total balls that we draw in Experiment # 2 Let W be the number of red balls that we draw in Experiment # 2 (e) Find the variance of Z. (In addition, find the standard deviation of Z.) (1) Find the variance of W. (In addition, find the standard deviation of W) (g) Find the covariance of Z and W (h) Interpret your answer to Part (g). (Is your answer what you would expect intuitively? Why?)

Explanation / Answer

Back-up Theory

Given 5 white, 3 black and 2 red balls, probability of a single draw giving a black ball = 3/10 = 0.3 ………………..(1)

and probability of a single draw giving a black or a white ball = (3 + 5)/10 = 0.8..........................................……(2)

If a discrete random variable, X, has probability function, p(x), then

Mean (average) of X = E(X) = sum{x.p(x)} summed over all possible values of x............................................... (3)

E(X2) = sum{(x2).p(x)} summed over all possible values of x…………..................................................………..(4)

Variance of X = V(X) = E(X2) – {E(X)}2……..…………………….......................................................……………..(5)

Given two random variables X and Y and their joint probability function, p(x, y),

Covariance of X and Y = Cov(X, Y) = E(XY) – {E(X) x E(Y)} ………..................................................….......……(6),

where E(XY) = sum{(xy).p(x,y)} summed over all possible values of x and y.................................................. ..(7).

Preparatory Work

Vide (1) and (2) of Back-up Theory, for X and Y as defined in the question,

P(X = 1) = 0.3 and hence P(X = 0) = 1 – 0.3 = 0.7

P(Y = 1) = 0.8 and hence P(Y = 0) = 1 – 0.8 = 0.2

E(X) = (1 x 0.3) + (0 x 0.7) = 0.3 …………………………………………………(8)

E(X2) = (12 x 0.3) + (02 x 0.7) = 0.3……………………………………………….(9)

E(Y) = (1 x 0.8) + (0 x 0.2) = 0.8 …………………………………………………(10)

E(Y2) = (12 x 0.8) + (02 x 0.2) = 0.8……………………………………………….(11)

Part (a)

Vide (5) of Back-up Theory, (8) and (9) above, V(X) = 0.3 – 0.32 = 0.21 ANSWER

Part (b)

Vide (5) of Back-up Theory, (10) and (11) above, V(Y) = 0.8 – 0.82 = 0.16 ANSWER

Part (c)

P(X = 0, Y = 0) = P(the ball is neither black nor white) = 1 – 0.8 = 0.2 ………….. (12)

P(X = 0, Y = 1) = P(the ball is white but not black) = 0.5……………. ………….. (13)

P(X = 1, Y = 1) = P(the ball is black) = 0.3………….........…………. ………….. (14)

(X = 1, Y = 0) does not make sense and hence its probability is zero.

Vide (7) of Back-up Theory,

E(XY) = (0 x 0 x 0.2) + (0 x 1 x 0.5) + (1 x 1 x 0.3) = 0.3.

Vide (6) of Back-up Theory,

Cov(X, Y) = 0.3 – (0.3 x 0.8) = 0.06 ANSWER

Part (d)

Apparently, X and Y look independent and hence one would expect the covariance to be zero.

But, a close look would reveal that X is contained in Y and not independent of Y. Hence covariance is not zero. ANSWER

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