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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