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# Probability Theory STAT 241

Yale

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This 4 page Class Notes was uploaded by Jeffrey Bogan on Thursday October 29, 2015. The Class Notes belongs to STAT 241 at Yale University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/231035/stat-241-yale-university in Statistics at Yale University.

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Date Created: 10/29/15

Chapter 10 Conditioning on a random variable With a contlnuous dlstrlbutlon At this point in the course I hope you understand the importance of the conditioning formula lEY info Ziwm infolEY Ft info for nite or countany in nite collections of disjoint events Fi for which S2 UiFi As a particular case if X is a random variable that takes only a discrete set of values 161 162 then E Y info 2 Ex x infolE Y t x xi info This formula can be simpli ed by the introduction of the function hot lEY X 16 info For then E Y info 2 PX xii infohx 1E hX info In this Chapter I want to persuade you that a similar formula applies when X has a continuous distribution with density function f given the info as lEY info lEhX info 00 hxfxdx 00 As a special case when Y equals the indicator function of an event B the formula reduces to PB OOPB X xfxdx From now on I will omit explicit mention of the conditioning information info writ ing hx for lEY X x There are several ways to arrive at formula as The most direct relies on the plausible assertion that EYX6Jhx ifJisasmall interval Wither The error of approximation should disappear as J shrinks to the point x Split R into a union of disjoint small intervals Ji xi xi Where 1641 x 6 then condition lEY 2 RX 6 JiEY x e Ji w E 5fxihxi w fa hxfxdx The combined errors of all the approximation should disappear in the limit as 6 tends to zero Statistics 241 25 October 2005 C101 David Pollard lt10lgt lt102gt Statistics 241 25 October 2005 Chapter 10 Conditioning on a random variable with a continuous distribution Alternatively we could start from a slightly less intuitive assumption that lEY should be nonnnegative if lEY l X x 3 0 for every x If we replace Y by Y 7 hX then we have EY7hX Xx lEYlXx7hx0 which gives lEY7hX 3 0 A similar argument applied to hX7Y gives lE hX 7 Y 3 0 Equality x follows REMARK Notice that formula also implies that E YgX E g at least for bounded functions g because lE Yg X l X x gxhx ln advanced probability theory the treatment of conditional expectations becomes most abstract Formula is used to de ne the conditional expectation hx lEY l X 6 One needs to show that there exists a random variable of the form hX which is uniquely determined up to trivial changes on sets of zero probability for which lEg X Y 7 hX 0 for every bounded g Essentially hX is the best approximation to Y using only information given by X With this abstract approach one then needs to show that conditional expecta7 tions have the properties that l have taken as axiomatic for Stat 241 Example lt101gt The convolution formula for densities derived from x The Poisson process is often used to model the arrivals of customers in a waiting line or the arrival of telephone calls at an exchange The underlying idea is that of a large popu7 lation of potential customers each of whom acts independently of all the others Example lt102gt A queuing problem with a surprising solution can be skipped EXAMPLES FOR CHAPTER 10 Example Suppose X and Y are independent random variables with continuous distri7 butions If X has density f and Y has density g then see Chapter 7 the random variable Z X Y has density 00 hltzgt 7 gltz7xgtfltxgtdx 700 The same formula can be derived from the formula 00 PB MB l X xfxdx 700 applied with B z E Z 3 z 6 for a small positive 6 Note that lP zZ5z6lXxlP z7xYz7x6lXx lP z7x Y z7x6 w6gz7x because X Y independent Invoke the conditioning formula Rz z s z 6 w w6gltz7xgtfltxgtdx 700 which leads us back to the convolution formula D Example Suppose an of ce receives two different types of inquiry persons who walk in off the street and persons who call by telephone Suppose the two types of arrival are described by independent Poisson processes with rate Aw for the walk7ins and rate As for David Pollard C1072 Chapter 10 Conditioning on a random variable with a continuous distribution the callers What is the distribution of the number of telephone calls received before the rst walkiin customer Write T for the arrival time of the rst walkiin and let N be the number of calls in 0 T The time T has a continuous distribution with the exponential density f l twe M for l gt 0 We need to calculate lP N i for i 01 2 From formula as with A equal to N i PNi AMPN139 t Ttftdl The conditional distribution of N is affected by the walkiin process only insofar as that pro cess determines the length of the time interval over which N counts Given T l the ran dom variable N has a Poissontcl conditional distribution Thus 00 7M A l i MN i Wkwe lw dl 0 1 Ali00 x i dx m ttt e u n x wt 0 ACHW ACMW p g w Lw lt AC gtliwxie xdx Aclw Aclw 1 0 The li and the last integral cancel Compare with lquoti 1 Writing 17 for twAC tw we have lP Niplipl fori0l2 That is l N has a geometricp distribution The random variable N has the distribution of the number of tails tossed before the rst head for independent tosses of a coin that lands heads with probability 17 Such a nice clean result couldn t happen just by accident Maybe we don t need all the Calculus to arrive at the distribution for N In fact the properties of the Poisson distribution and Problem 81 show what is going on as I will now explain Consider the process of all inquiries both walkiins and calls In an interval of length l the total number of inquiries is the sum of a Poissontwl distributed random variable and an independent Poissontcl distributed random variable the total has a Poissontwl Lcl distribution Both walkiins and calls contribute independent counts to disjoint intervals the total counts for disjoint intervals are independent random variables It follows that the pro cess of all arrivals is a Poisson process with rate tw LC Now consider an interval of length l in which there are X walkiins and Y calls From Problem 81 given that X Y n the conditional distribution of X is Binn p where 7 th 7 Aw 7 th Acz 7 Aw AC That is X has the conditional distribution that would be generated by the following mecha7 nism P 1 Generate inquiries as a Poisson process with rate tw LC 2 For each inquiry toss a coin that lands heads with probability p tw tw AU For a head declare the arrival to be a walkiin for a tail declare it to be a call A formal proof that this twoistep mechanism does generate a pair of independent Poisi son processes with rates AW and LC would involve 1 Prove independence between disjoint intervals Easy 2 If step 2 generates X walkiins and Y calls in an interval of length I show that lP X iYj lP XilP Yj X N Poissontwl and Y N Poissontcl Statistics 241 25 October 2005 C103 David Pollard Chapter 10 Conditioning on a random Variable With a continuous distribution You should be able to write out the necessary conditioning argument for 2 The twoistep mechanism explains the appearance of the geometric distribution in the problem posed at the start of the Example The classi cation of each inquiry as either a walkiin or a call is effectively carried out by a sequence of independent coin tosses with probability p of a head a walkiin The problem asks for the distribution of the number of tails before the rst head The embedding of the inquiries into continuous time is irrelei vant D Statistics 241 25 October 2005 C104 David Pollard

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