linearity of conditional expectation

/Filter /FlateDecode The possible values of \(\textrm{E}(X|Y)\) are determined by \(\textrm{E}(X|Y=y)\) for each possible value \(y\) of \(Y\), and the corresponding probabilities are determined by the distribution of \(Y\). For a given value \(x\) of \(X\), \(\textrm{E}(Y|X=x)\) is a number. The best answers are voted up and rise to the top, Not the answer you're looking for? Use the table above. Let and be integrable random variables, 0 and c, c1, c2 be real numbers. Example 5.43 Continuing Example 5.40. \textrm{E}(X|Y=1) & = (-1)(25/40) + (0)(0)+(1)(15/40) = -1/4\\ 3 0 obj << Theorem 5.5 For any random variables \(X\) and \(Y\). Conditioningar.vbyadiscreter.v Example4:WheneverX andY arediscreterandomvariables,ComputationofE[X|Y] canbehandledasinexample3. 1.1 Law of Iterated Expectations. (a) Linearity. The regression problem. Conditional Expectation Example: Suppose X,Y iid Exp().Note that FY (a x) = 1 e(ax) if a x 0 and x 0 (i.e., 0 x a) 0 if otherwise Pr(X + Y < a) = Z < FY (a x)fX(x)dx Z a 0 (1 e(ax))ex dx = 1 ea aea, if a 0. d da Pr(X + Y < a) = 2aea, a 0. To achieve the pattern HT, we first need to flip until we get H, and then we complete the pattern once we get the first T after that. \textrm{E}(\textrm{E}(X|Y)) = \int_2^{6.5} w \left((8/81)(w-2)\right)dw = 5 But when trying for HT, any H that follows H just maintains our current position. Various types of "conditioning" characterize some of the more important random sequences and processes. \], \(\textrm{E}(X^2) = \textrm{Var}(X) + (\textrm{E}(X))^2 = 1/12 + (1/2)^2=1/3\), \[ Let X and Y be discrete random variables. 2/63 10.3 Properties of Conditional Expectation It's helpful to think of E(jG ) as an operator on random variables that transforms F-measurable variables into G-measurable ones. \], \(\textrm{E}(U_1|Y) = 0.5\textrm{E}(X|Y) = 0.5(1.5Y + 0.5)=0.75Y + 0.25\), \(\textrm{E}(U_1|Y) = 0.5Y + 0.5(Y+1)/2 = 0.75Y + 0.25\), \[ Deriving the OLS formula as a means of approximating the conditional expectation function. Theorem 2.3.2 (FUndamental properties of conditional expecta tions). \]. Given a value \(x\) of \(X\), the conditional expected value \(\textrm{E}(Y|X=x)\) is a number. If JWT tokens are stateless how does the auth server know a token is revoked? This is true for any value \(y\) of \(Y\). \end{align*}\], Differentiate both sides with respect to \(w\); remember the chain rule. general linear regression model assumptionslife celebration memorial powerpoint template. The minimizing value of Z is the conditional expected value of X. Theorem 115 (conditional expectation as a projection) Let G F be sigma-algebras and X a random variable on (,F,P). The height \(Y\) is a random variable whose conditional distribution given \(X=x\) is Uniform(0, \(x\)). x\Ks7WpodU8#qvJI%9H%F$U9X`ntM#]L~?s}5+gs>n~u6bLxt% 7\FM3f6L X0 g/]}>vUm!F5F43F}9_9" Compute the expected value using this conditional distribution: \(\textrm{E}(X|Y=4) = 5(2/7) + 6(2/7) +7(2/7) + 8(1/7)= 6.29\). The expected value of a random variable with a finite number of outcomes is a . When trying for HH, any T that follows H destroys our progress and takes us back to the beginning. Since each of the conditional expectations E(U jY) and E(V jY) is a function of Y, so is the linear combination aE(U jY)bE(V jY). &= \left( \sum_{i=1}^k a_i~E(X_i \mid Y=y) \right) + a_{k+1}~E(X_{k+1} \mid Y=y) \\ My professor says I would not graduate my PhD, although I fulfilled all the requirements. Proof that linearity of expectation holds for countably infinite sum of random variables $(X_n)$ given $\sum_{i=1}^{\infty}E[|X_i|]$ converges? \], \(\textrm{E}(XY|X=0.5)=\textrm{E}(0.5Y|X=0.5) = 0.5\textrm{E}(Y|X=0.5) = 0.5(0.25)=0.125\), \(\textrm{E}(XY|X=0.2)=\textrm{E}(0.2Y|X=0.2) = 0.2\textrm{E}(Y|X=0.2) = 0.2(0.1) = 0.02\), \(\textrm{E}(XY|X=x)=\textrm{E}(xY|X=x) = x\textrm{E}(Y|X=x)\), \(\textrm{E}(XY|X=x)=x\textrm{E}(Y|X=x)\), \(\textrm{E}(XY|X=x)=x\textrm{E}(Y|X=x)=x(x/2)=x^2/2\), \[ 7.1. Thus, we have . Making statements based on opinion; back them up with references or personal experience. The following is a simulation of the lookaway challenge problem. If the linear model is true, i.e., if the conditional expectation of Y given X indeed is a linear function of the X j 's, and Y is the sum of that linear function and an independent Gaussian noise, we have the following properties for least squares estimation. 2. The table below defines the function \(\ell\). conditional mean), of a random variable \(Y\) given the event \(\{X=x\}\), defined on a probability space with measure \(\textrm{P}\), is a number denoted \(\textrm{E}(Y|X=x)\) representing the probability-weighted average value of \(Y\), where the weights are determined by the conditional distribution of \(Y\) given \(X=x\). By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. What to throw money at when trying to level up your biking from an older, generic bicycle? In the previous example, suppose that every day when Regina arrives she guesses what time Cady arrived/will arrive that day (if Regina arrives second/first). Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, I started doing it by induction up there ^. On the last step I separated the $(k+1)^{\text{th}}$ "term" since I'm trying to find a way to use the induction hypothesis but I need to do something to get rid of the $(k+1)^{\text{th}}$ integral, as well the $(k+1)^{\text{th}}$ random variable in the underlying conditional distribution Following is a list of some of the more important of these. The dashed line is a linear regression model for these two variables. Can you explain why? &+a_{k+1}\int_{-\infty}^{\infty}x_{k+1}f_{X_{k+1}|Y}(x_{k+1}|y)~dx_{k+1}\\ The random variable \(\textrm{E}(Y|X)\) is a function of \(X\). @Did In this problem, the hard part is doing the base case, and then the induction step is to simply use the base case by grouping the rest of the $k$ variables as another variable, say $Z$ and applying the base case to pull out 1 variable, and then applying the induction hypothesis on $Z$. (Count the tosses that result in H, so H is 1 flip, TH is 2 flips, TTH is 3 flips, etc). Compute and interpret the expected number of rounds in a game. For each \(2

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