12_DiscreteRVs-4_Joint_PMFs_packed

Conmnued 0 120 120 3 120 220

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: probability 1 pZ(3) = P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4) = P(X+2Y = 4) = P(X=2 and Y=1) = 1/20 pZ(5) = P(X+2Y = 5) = P(X=1 and Y=2) + P(X=3 and Y=1) = 2/20, etc Ilya Pollak Example 2.9 (conMnued) 0 1/20 1/20 3 1/20 2/20 3/20 2 1/20 2/20 3/20 1 1/20 1/20 1/20 0 1 2 3 4 x 1/20 2/20 2/20 4/20 3/20 9 1/20 4 8 1/20 1/20 7 1/20 3 3/20 10 2/20 11 1/20 12 4 pZ(k) 6 Joint PMF pX,Y(x,y) y k 5 Z=X+2Y (a)  Find pZ(k). (b)  Find E[Z]. 1/20 E[Z] = 3 · (1/20) + 4 · (1/20) + 5 · (2/20) + 6 · (2/20) + 7 · (4/20) + 8 · (3/20) + 9 · (3/20) + 10 · (2/20) + 11 · (1/20) + 12 · (1/20) = 7.55 Ilya Pollak Example 2.9 (conMnued) Z=X+2Y Joint PMF pX,Y(x,y) y (b) Find E[Z]. 4 0 1/20 1/20 1/20 3/20 3 1/20 2/20 3/20 1/20 7/20 2 1/20 2/20 3/20 1/20 7/20 1 1/20 1/20 1/20 0 3/20 1 2 3/20 6/20 3 8/20 4 x 3/20 Alternatively, E[Z] = E[X] + 2E[Y] = 51/20 + 2 · (50/20) = 7.55 E[X] = 1 · (3/20) + 2 · (6/20) + 3 · (8/20) + 4 · (3/20) = 51/20 E[Y] = 1 · (3/20) + 2 · (7/20) + 3 · (7/20) + 4 · (3/20) = 50/20 Ilya Pollak Example 2.10: mean of a binomial random variable •  •  •  •  •  N students in class Each has probability p of ge\ng an A All grades are independent X = number of students that get an A What is E[X]? Ilya Pollak Example 2.10: mean of a binomial random variable ⎧ 1, if the i-th student gets an A ⎪ Let Xi = ⎨ ⎪ 0, otherwise ⎩ X1 , X2 ,…, X N are Bernoulli random variables with E[ Xi ] = p. X1 + X2 + … + X N =X is the number of successes in N independent Bernoulli trials. Therefore, X is a binomial random variable, with the following PMF: ⎛N⎞ k pX (k ) = ⎜ p (1 − p )N − k , for k = 0,… N ⎝k⎟ ⎠ Hard way of evaluating E[X ]: ⎛N⎞ k E[X ]=∑ k ⎜ p (1 − p )N − k ⎟ k=0 ⎝ k ⎠ N Easy way: E[X ]=E[X1 ]+E[X2 ]+… +E[X N ] = Np E.g., if N =60 and p =1/5, then E[X ]=12. Ilya Pollak Example 2.11 •  n people throw their hats in a box •  each picks up one hat at random •  X = number of people who get their own hat back •  What is E[X]? Ilya Pollak Example 2.11 ⎧1, if the i - th person gets his own hat Let X i = ⎨ ⎩0, otherwise ⎧1 / n, if k = 1 ⎪ pX i ( k ) = ⎨1 − 1 / n, if k = 0 ⎪ 0, otherwise ⎩ ⎛ 1⎞ 1 1 E [ X i ] = 1 ⋅ + 0 ⋅ ⎜1 − ⎟ = ⎝ n⎠ n n X = X1 + X 2 + … + X n 1 E [ X ] = E [ X1 ] + E [ X 2 ] + … + E [ X n ] = n ⋅ = 1 n Ilya Pollak CondiMoning a random variable on an event •  CondiMonal PMF of a random variable X, condiMoned on an event A with P(A)>0: P({ X = x} ∩ A) pX | A ( x ) = P( X = x | A) = P( A) Note : P( A) = ∑ P({ X = x} ∩ A), and therefore ∑p x X |A ( x) =...
View Full Document

This note was uploaded on 09/11/2013 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue University.

Ask a homework question - tutors are online