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Tutorial 8, STAT1301 Fall 2010, 16NOV2010, MB103@HKU By Joseph Dong Recall: A Partition on a Set ο ο Any exhaustive and disjoint collection of subsets of a given set forms a partition of that set. ο E.g. ο π΅, π΅π forms a trivial partition of the presumed set π = π΅ βͺ π΅π . ο If π: π β π = 1,2,3 , then the collection of pre-images of atoms of the range, π β1 1 , π β1 2 , π β1 3 , forms a partition of the domain π. 2 Recall: Conditioning on a Partition ο ο Shares the same idea with ο Divide and Conquer ο Casewise enumeration ο A Tree-diagram ο Formal language: ο Goal = find the probability of event πΈ, β πΈ . ο It is equivalent to finding the intersection of it with the sure event Ξ©. β πΈ β‘β πΈβ©Ξ© . 3 Recall: Conditioning on a Partition (contβd) ο ο Formal language (continued) ο Now break down the sure event into a number of manageable smaller pieces and these pieces together forms a partition {π΄π |π β πΌ} of the sure event Ξ©. ο If we investigate all such events πΈ β© π΄π , then weβre done. β πΈ β© Ξ© = ββ πΈ β© π΄π ο The hardcore of the problem now becomes finding each β πΈ β© π΄π , and this is where the conditioning takes place. β πΈ π΄π β β π΄π ο Assuming it is more straight forward a task to find β πΈ π΄π and β π΄π . 4 Recall: What does an R.V. do to its State Space? ο An r.v. cuts the state space into blocks. On each of these blocks, the r.v. sends all points there to a common atom in the sample space. ο ο An r.v. causes a partition on the state space. ο Conversely, given a partition on the state space, you can also define random variables on it so that it βconformsβ the partition by taking one value for each block. Random Variable Partition on π 5 Conditioning an Event on an R.V. ο ο Since an r.v. cuts the state space into a partition, conditioning on an r.v. is just conditioning on that partition it caused on the state space. ο The meaning of β πΈ π is now clearly illustrated on the right. 6 β πΈ π as a Random Variable ο ο It contains a random variable π inside, making itself a function of π. ο It has a distribution and expectation. ο Lotus ο Question: Whatβs the meaning of its expected value? ο To fix its value by fixing an π value: ο β πΈ π = π₯1 , β πΈ π β π₯1 , π₯2 ο Every fixed value is now a conditional probability involving two events. 7 Exercise: Finding β πΈ from β πΈ π ο ο This is the prototypical problem of finding the probability of an event via the technique of conditioning on a random variable. ο Hint: Ponder on the link between Law of Total Probability and Expectation. ο Ans: β πΈ =πΌβ πΈπ 8 β ππ ο ο It involves two r.v.βs now. ο Given β π π : β π π is a function of the ο Q1: How to find β π = bivariate random vector π, π . ο Fixing π will give you back the conditional density of π given π at the fixed position. 9 Conditional, Marginal, and Joint densities ο ο Difference among 3 types of densities: ο a conditional density β π π ο is normalized by the marginal probability of β π ο is a point dividing a row sum/integral ο is the density of π|π ο a joint density β π, π ο is normalized by the entire joint space ο is a point dividing the sum/integral of entire space ο is the density of π, π ο a marginal density β π ο is also normalized by the entire space ο is a row sum dividing the sum/integral of entire space ο is the density of π 10 Handout Problem 1 ο 11 Recall: Whatβs the Expectation of a random variable ο ο First of all, the random variable has to be numerically valued. Thatβs why expectation is also known as the βexpected valueβ and is a numerical characteristic of the sample space (a subset of β or simply β itself with zero densities equipped at those impossible points). +β πΌ π = π₯ππ π₯ ππ₯ ββ ο The expectation is both conceptually and technically equivalent to the location of the center of probability mass of the sample space. ο Expectation provides only partial information of the random variable because it eliminates randomness by giving you back only 1 representative point of the sample space. 12 For examples, ο ο πΈ|π is a set-valued random variable. ο Given π = π₯, it evaluates to the set πΈ β© π = π₯ . ο We cannot have an expected value defined for πΈ|π. ο Clarification: πΈ|π is not β πΈ π . The latter is numerically valued, as we have previously established for its expected value: πΌ β πΈ π = β πΈ . ο More elaboration: On the set-theory layer, πΈ|π is not strictly different from the set-r.v. pair πΈ, π . But when onto the probability-theory layer, β πΈ π is normalized by a different space than is β πΈ, π . 13 πΌ ππΈ ο ο π|πΈ is a numerically-valued random variable. We can compute its expected value. ο πΌ π πΈ vs πΌ π : their sample spaces are different. ο Compute πΌ π πΈ using βπΈ = β β πΈ β π π Ξ© β ππ, πΈ β πΈ +β = β β ,πΈ) β πΈ π₯ππ|πΈ π₯ ππ₯ ββ ο Compute πΌ π using β +β π π β ππ = Ξ© π₯ππ π₯ ππ₯ ββ 14 Warm-up exercise ο ο Handout Problem 2 15 πΌ π π : concepts ο ο First of all, this is a random variableβa function of π. ο Its randomness comes from the state space of π, but the mapping mechanism is worked out together by both of π and π. ο This expression is known as the conditional expectation of the conditionee π given the conditioner π. ο The expectation is done with respect to π. ο To be precise, should say w.r.t. π|π. ο There are multiple (or even a continuum of) sample spaces of π|π, depending on which atom value π takes. After fixing π to an atom, or equivalently, a block in the state space that has been partitioned by π, the expression πΌ π π = π₯1 is just a constant. ο The expectation eliminates the randomness of π given π. 16 πΌ π π as an r.v. ο ο It uses the joint state space of π and π as its own state space. ο It uses a degenerated version of the sample space of π as its own sample space. ο The degeneration preserves the locus of the overall center of mass. ο Each point in the degenerated space is a block center of mass 17 βDegeneration preserves overall center of massβ ο ο π cuts its own state space as well as the joint state space of it and π. ο This partition of the joint state space will be mapped by π to a partition on its own sample space (a numeral set). ο Then the expression πΌ π π = π₯1 represents the locus of center of mass of the first block of the partition. ο πΌ π π represents the totality of loci of these block centers of mass. 18 Exercise: Finding πΌ π from πΌ π π ο ο This is the prototypical problem of finding the expectation of a random variable via the technique of conditioning on another random variable. ο Ans. πΌ π =πΌ πΌ ππ ο In the divide-conquer-merge paradigm: ο Divide is done by the conditioner π ο Conquer refers to the inner expectation carried out at each division ο Merge refers to the outer expectation to piece up the whole plate. This exercise addresses the merge step. ο Compare with the conditional probability, ponder the link between them. 19 Conditional Variance ο ο Finding variance by conditioning: π π =πΌ π π π +π πΌ π π ο Pf. ο Unfortunately, the degeneration of the sample space of π does not preserve second moments. ο Thatβs why there is the addendum π πΌ π π in the formula. 20 Summary: Conditional Expectation ο The key observations are ο Obs1: To find the center of mass of a piece of material, you can divide it into a few blocks, find their centers of mass, and then find the center of mass of these block centers of masses. The initial division of the piece is quite arbitrary. ο This fundamental law of physics supports the many nice properties of expectation in the calculus of probability. ο Obs2: A random variable partitions its state space into a collection of atom-valued blocks. ο This suggests using random variable as a general device to divide the piece mentioned in Obs1. Such a random variable is called the conditioner. 21 Linking β πΈ π to πΌ π π ο ο Trick: Use indicator of set πΈ. The indicator is a Bernoulli random variable. ο Reason: β πΈ π β‘ πΌ πΌπΈ π ο Conclusion: The conditional probability of an event conditioned on a random variable (a partition) is a conditional expectation of the indicator of that event conditioned on the same random variable in disguise. ο All properties of conditional expectation should apply to conditional probability. Such as the Law of Total Probability is just πΌ π β‘ πΌ πΌ π π in disguise. 22 Choosing Conditioner ο ο The art of conditioning lies in the choice of the conditioner. ο Usually, if our unknown target is the r.v. π, and we know that π is a known function of a known r.v. π, then it would be natural to use π as the conditioner for π, that is ο Divide the state space of π by π ο Conquer every πΌ π π ο Merge them into πΌ π 23 Exercises ο ο Handout problem 3 ο Handout problem 4 ο Handout problem 5 ο Handout problem 6 24