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# slides9 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 9 AD Feb. 2010 AD () Feb. 2010 1 / 13

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Example: Optimal Stock Vancouver 2010 Olympic Torchbearer red mittens are currently sold at a net pro°t of b \$ for each unit sold and will bring a net loss of l \$ for each unit left unsold after the Olympics. The number of units of the product sold at a speci°c store is a random variable X with pmf p ( i ) , i ° 0. If the store must stock this product in advance, compute the expected pro°t for a determined number of units s . Find b s which optimizes the maximum expected pro°t. The pro°t is equal to Y ( s ) = bX ± l ( s ± X ) if X ² s = sb if X > s Hence the expected pro°t is E [ Y ( s )] = s i = 0 ( bi ± l ( s ± i )) p ( i ) + i = s + 1 sbp ( i ) = sb + ( b + l ) s i = 0 ( i ± s ) p ( i ) AD () Feb. 2010 2 / 13
Example: Optimal Stock We want to maximize the pro°t. We note that E [ Y ( s + 1 )] ± E [ Y ( s )] = b ± ( b + l ) s i = 0 p ( i ) so we have E [ Y ( s + 1 )] ± E [ Y ( s )] ° 0 is equivalent to s i = 0 p ( i ) < b b + l . So we should consider s ³ the largest value such that s ³ i = 0 p ( i ) < b b + l then E ( Y ( 0 )) < ´ ´ ´ < E ( Y ( s ³ )) < E ( Y ( s ³ + 1 )) > E ( Y ( s ³ + 2 )) > ... so the optimal stock is b s = s ³ + 1. AD () Feb. 2010 3 / 13

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Variance A r.v. X is entirely de°ned by its p.m.f. but it is very useful to be able to summarize the essential properties of it through a few suitably de°ned measures. We have de°ned previously the expected value E ( X ) . However this measure tells us nothing about the variations of X ; e.g. consider X = 0 with proba. 1, Y = 1 w.p. 0.5 and Y = ± 1 w.p. 0.5 and °nally Z = 100 w.p. and Z = ± 100 w.p. 0.5 then E ( X ) = E ( Y ) = E ( Z ) = 0 whereas there is a much greater spread/variability in the values of Y
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slides9 - Lecture Stat 302 Introduction to Probability...

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