10_DiscreteRVs-2_Expectation_packed

# Mean square deviaon of the esmate of p

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Unformatted text preview: , whereas the second one is riskless. •  Standard devia:on characterizes the risk: –  Opportunity 1: st.dev. = (0.5(500+1000)2 + 0.5(500−2000)2)1/2 = 1500 –  Opportunity 2: st.dev. = (1(500−500)2)1/2 = 0 •  Standard devia:on characterizes the spread of possible proﬁts around the expected proﬁt. Ilya Pollak Problem 2.20: Expecta:on and variance of a geometric random variable As an ad campaign, a chocolate factory places golden :ckets in some of its candy bars, with the promise that a golden :cket is worth a trip through the chocolate factory, and all the chocolate you can eat for life. If the probability of ﬁnding a gold :cket is p, ﬁnd the mean and the variance of the number of candy bars you need to eat to ﬁnd a :cket. Ilya Pollak Mean of a geometric random variable •  Let C = # candy bars un:l 1st success •  Model C as geometric with parameter p: ⎧(1 p) k −1 p, k = 1, 2,… pC ( k ) = ⎨ 0, otherwise ⎩ ∞ E [C ] = ∑ k (1 − p) k −1 p k =1 Useful fact : k (1 − p) k −1 d = − {(1 − p) k } dp Ilya Pollak Mean of a geometric random variable •  Let C = # candy bars un:l 1st success •  Model C as geometric with parameter p: ⎧(1 − p) k −1 p, k = 1, 2,… pC ( k ) = ⎨ 0, otherwise ⎩ ∞ E [C ] = ∑ k (1 − p) k −1 p k =1 Useful fact : k (1 − p) k −1 d = − {(1 − p) k } dp Ilya Pollak Mean of a geometric random variable •  Let C = # candy bars un:l 1st success •  Model C as geometric with parameter p: ⎧(1 − p) k −1 p, k = 1, 2,… pC ( k ) = ⎨ 0, otherwise ⎩ ∞ E [C ] = ∑ k (1 − p) k −1 p k =1 d Useful fact : k (1 − p) = − {(1 − p) k } dp ∞ ⎫ d d ⎧∞ k k Therefore, E [C ] = − p∑ {(1 − p) } = − p ⎨∑ (1 − p) ⎬ dp ⎩ k =1 ⎭ k =1 dp k −1 ⎧ 1⎫ 1 d ⎧1 − p ⎫ = −p ⎨ ⎬ = − p⎨− 2 ⎬ = dp ⎩ p ⎭ ⎩ p⎭ p Ilya Pollak How many candy bars un:l ﬁrst success? •  If there are ﬁve golden :ckets per 1,000,000 bars, then p=1/200,000. •  [Note: we assume an inﬁnite number of bars so that C is truly geometric.] •  Then E[C] = 200,000. •  On average, have to buy 200,000 chocolate bars un:l ﬁrst success. Ilya Pollak E[C2] ∞ E [C 2 ] = ∑ k 2 (1 − p) k −1 p k =1 Another useful fact : k (1 − p) 2 k −1 d2 d k +1 = 2 {(1 − p) } + {(1 − p) k } dp dp [Because the right - hand side is ( k + 1) k (1 − p) k −1 − k (1 − p) k −1 .] d2 ⎧ ∞ d ⎧∞ 2 k +1 ⎫ k⎫ E ⎡C ⎤ =p 2 ⎨∑ (1 − p ) ⎬ + p ⎨∑ (1 − p ) ⎬ ⎣ ⎦ dp dp ⎩ k 1 ⎩ k =1 ⎭ = ⎭ (1− p )2 / p −1 / p ⎫1 ⎫ 1 2p 1 2 1 d2 ⎧ 1 d⎧ 1 = p 2 ⎨ − 2 + p⎬ − = p ⎨− 2 + 1⎬ − = 3 − = 2 − dp ⎩ p dp ⎩ p pp p ⎭p ⎭pp Ilya Pollak Variance of a geometric random variable 2 1 ⎛1⎞ 11 2 2 var(C ) = E [C ] − ( E [C ]) = 2 − − ⎜ ⎟ = 2 − p p ⎝ p⎠ p p 2 If p = 1/200, 000, the standard deviation is 200, 000 2 − 200, 000 ≈ 200, 000 Thus, your actual number candy bars until first success may be quite far from the mean! Ilya Pollak...
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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.

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