Question

(a) Consider four independent rolls of a 6-sided die. Let X be the number of l's and let y be the number of 2's obtained. What is the joint PMF of X and Y? (b) Let X1, X2, X3 be independent random variables, uniformly distributed on [0,1]. Let Y be the median of X1, X2, X3 (that is the middle of the three values). Find the conditional CDF of X1, given that Y = 0.5. Under this conditional distribution, is X1 continuous? Discrete?

Answer 1

SOLUTION :

From given data :

• a) probability of getting a 1 =1/6 and probability of getting a 2 =1/6

  joint pmf of X and Y is as follows:

  P(X=x;Y=y)=   $\frac{4!}{x!*y!*(4-x-y)!}*(1/6)^{x}(1/6)^{y}(4/6)^{4-x-y}$

  hence :

(x,y) P(X,Y) (X=0,Y=0) 0.1975 (X=1,Y=0) (X=2,Y=0) (X=3,Y=0) (X=4,Y=0) (X=0,Y=1) 0.1975 0.0741 0.0123 0.0008 0.1975 (X=1,Y=1) (X=2,Y=1) (X=3,Y=1) (X=0, Y=2) 0.1481 0.0370 0.0031 0.0741 (X=1,Y=2) 0.0370 (X=2, Y=2) 0.0046 0.0123 (X=0, Y=3) (X=1, Y=3) 0.0031 (X=0,Y=4) 0.0008

• b) There are three possibilities ;

X₁ is either the smallest,the median,or the largest.each of these possibilities occurs with probability 1/3.

If X₁ is the smallest ,then given that the median is 1/2.its distribution function is uniform in (0,1/2).

If X₁ is the median it is equal to 1/2 with probability 1.

If X₁ is the largest then conditioned on the median being 1/2 it is uniform in (1/2,1).

HENCE , P(X₁< x ) = 2x/3 if x <1/2;2/3 if x =1/2;and 2/3+(1/3)(x-1/2)/(1/2) if x>1/2 .

For conditioned on Y = 1/2 we have that X1 is neither continous nor discrete .It has a jump on x =1/2,so that it cannot have a density ,and thus it is not continous ;and there are infinitely many values it can take ,so that it is not discrete.