12_DiscreteRVs-4_Joint_PMFs_packed

Conmnued 0 120 120 3 120 220

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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) =...
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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.

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