# 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 ± 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 ² the largest value such that s ² i = 0 p ( i ) < b b + l then E ( Y ( 0 )) < ³³³ < E ( Y ( s ² )) < E ( Y ( s ² + 1 )) > E ( Y ( s ² + 2 )) > ... so the optimal stock is b s = s ² + 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
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## This note was uploaded on 10/21/2010 for the course STAT Stat302 taught by Professor 222 during the Spring '10 term at The University of British Columbia.

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

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