12_DiscreteRVs-4_Joint_PMFs_packed

Up one hat at random x number

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Unformatted text preview: 1 x Ilya Pollak CondiMoning a random variable on another random variable •  If X and Y are random variables, condiMonal PMF pX|Y of X given Y is defined as follows, for all y such that pY(y) >0: pX |Y ( x | y ) = P( X = x | Y = y ) = P({ X = x} ∩ {Y = y}) P({Y = y}) pX ,Y ( x, y ) = pY ( y ) Ilya Pollak CondiMonal PMF: examples Joint PMF pX,Y(x,y) y x 2 3 4 4 0 1/20 1/20 1/20 pX|Y(x|4) 1/3 1/3 1/3 3 1/20 2/20 3/20 1/20 x 2 1/20 2/20 3/20 1/20 pX|Y(x|3) 1/7 2/7 3/7 1/7 1 1/20 1/20 1/20 0 1 2 4 3 1 2 3 4 x y pY|X(y|3) 1 1/8 2 3/8 3 3/8 4 1/8 Ilya Pollak Example: memoryless property of a geometric random variable •  X = the number of independent coin flips unMl first H, with P(H) = p. ⎧(1 − p) k −1 p, if k = 1, 2,… pX ( k ) = ⎨ ⎩ 0, otherwise •  Given that the first trial is a T, what is the condiMonal PMF of X – 1 ? Y = X −1 A = { X ≥ 2} pY | A ( m) = ? •  IntuiMon: since flips are independent, the outcome of the first flip does not influence the future. Therefore, the remaining flips aher the first one should have the same condiMonal distribuMon as pX. In other words, the answer must be pY|A = pX. Ilya Pollak Example: memoryless property of a geometric random variable •  X = the number of independent coin flips unMl first H, with P(H) = p. ⎧(1 − p) k −1 p, if k = 1, 2,… pX ( k ) = ⎨ ⎩ 0, otherwise •  Given that the first trial is a T, what is the condiMonal PMF of X – 1 ? Y = X − 1; A = { X ≥ 2} ⎧ P( X = m + 1) P(Y = m, X ≥ 2) P( X = m + 1, X ≥ 2) ⎪ , if m = 1, 2,… pY | A ( m) = = = ⎨ P( X ≥ 2) P( X ≥ 2) P( X ≥ 2) ⎪0, otherwise ⎩ ⎧ (1 − p) m p ⎧ (1 − p) m −1 p, if m = 1, 2,… ⎪ , if m = 1, 2,… =⎨ =⎨ = pX ( m) 1− p ⎩0, otherwise ⎪0, otherwise ⎩ Ilya Pollak Example: memoryless property of a geometric random variable •  X = the number of independent coin flips unMl first H, with P(H) = p. ⎧(1 − p) k −1 p, if k = 1, 2,… pX ( k ) = ⎨ ⎩ 0, otherwise •  Given that the first N trials are all T’s, find condiMonal PMF of X – N. YN = X − N ; AN = { X ≥ N + 1} ⎧ (1 − p) m −1 p, if m = 1, 2,… pYN | A N ( m) = ⎨ = pX ( m) ⎩0, otherwise Ilya Pollak Example s Joint PMF pR,S(r,s) 3 6/45 9/45 0 2 6/45 9/45 3/45 1 4/45 6/45 2/45 2 3 Let A = {S ≠ 3} (a)  Find pS(s) and pS|A(s). 1 r Ilya Pollak Example s Joint PMF pR,S(r,s) 3 6/45 9/45 0 2 6/45 9/45 3/45 1...
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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-West Lafayette.

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