linearity of conditional expectation proof

\end{array}. We use the de nition, calculate and obtain E[X] = 2 1 36 + 3 1 36 + + 12 1 36 = 7: As stated already, linearity of expectation allows us to compute the expected value of a sum of random variables by computing the sum of the individual expectations. The red arrows represent Properties of conditional expectation. way to calculate it using linearity. name. Let X and Y be two independent G e o m e t r i c ( p) random variables. 7.1.2 Some properties of conditional expectation. The lemma below shows that practically all properties valid for usual (complete) mathematical expectation remain valid for conditional expectations. \end{equation}\], \[\begin{align*} Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. \] Thus, we have . 14.1: Conditional Expectation, Regression - Statistics LibreTexts that the joint distribution of \(X\) and \(Y\) is \(f(x, y)\). But linearity of expectation says that to calculate \(E[Y - X]\), it does not matter how \(X\) and \(Y\) are &= \sum_x x f(x) + b\sum_x f(x) & \text{(factor constant outside the sum)} \\ Find the expected number of locations with no phone numbers stored, the 4. given another r.v., and discuss key properties such as taking out what's known, Adam'. Chapter 1 Expectation Theorems | 10 Fundamental Theorems for - Bookdown As I explained, my understanding of the proof leads me to blatantly problematic statement. Each time you buy a Happy Meal, you are equally Hint: Express this complicated random variable as a sum of geometric random variables, and /Font << /F53 4 0 R /F55 5 0 R /F56 6 0 R /F26 7 0 R /F29 8 0 R /F24 9 0 R /F15 10 0 R /F38 11 0 R /F30 12 0 R >> \], \[ X(X-1) = \sum_{i=1}^n \sum_{j\neq i} Y_{ij}. 5.5 Expected values of linear combinations of random variables By definition, the conditional mean of $Y$ on $X$ is a random variable $\psi$ &= E[E[Y| N] + \dots + E[Y| N]|N] \\ =&\arg \min_{g(x)} E \Big[ 2 \big(Y - E(Y|X)\big) \big(E(Y|X) - g(X)\big) + \big(E(Y|X) - g(X)\big)^2\Big]\\ PDF Lecture 10 : Conditional Expectation - University of California, Berkeley /Filter /FlateDecode \end{align*}. for fast information retrieval. \(n=4\) and \(X=3\). $$\Rightarrow E(Y^2\mid X) -2E\Big [(YE(Y \mid X))\mid X\Big] + E\Big [(E(Y \mid X))^2\mid X\Big] \\\le E(Y^2\mid X) -2E\Big [(Yh(X))\mid X\Big] + E\Big [(h(X))^2\mid X\Big]$$, The first term of the LHS and the RHS cancel out. It would be great if you can elaborate on your answer. Its simplest form says that the expected value of a sum of random variables is the sum of the expected values of the variables. Each member of the group draws a name at random from the hat and must by a gift for that person. Let \(Y_{ij}\) be 1 if tickets \(i\) and \(j\) 1. I(]SE6%Uyuc2\TXMc(5mSl|nSgN9_LQtWgH'{sgg`307!bw@;Qs7{H%i5Gi&MQ]nYy43,4sP2m{a}8wx~510N6 H.}i-nA#EZ*m(/cvkd}h69j9>T0O31jsr|}N ^S**xQlm4/uc'{PY%pa'WW"Y,Gw-D eNa8P 1IMWom:Y;WY eIzC3#&()6C/d9H Because of the minus sign in front of $\big(Y - h(X) \big) \big(h(X) - g(X)\big)$ one could find some $g(X)$ such that $E \Big[ - 2 \big(Y - h(X) \big) \big(h(X) - g(X)\big) + \big(h(X) - g(X)\big)^2\Big] < 0$. Next, recall that X lim "X ^n. 4.7: Conditional Expected Value - Statistics LibreTexts \] $E \Big[ 2 \big(Y - E(Y|X)\big) \big(E(Y|X) - g(X)\big) + \big(E(Y|X) - g(X)\big)^2\Big] = E( 0 + 0)$ = 0. . In the diagram below, is (This actually characterizes $E(Y|X)$, if one inspects the proof of existence). The conditional expectation always exists. \], \[ E[Y_i] = 0 \cdot \frac{N_0}{N} + 1 \cdot \frac{N_1}{N} = \frac{N_1}{N}. I've seen both E ( Y | X) and E ( Y | X = x) referred to as the "conditional expectation function". k + 1 integrals(a1x1 +. \]. Denition 16.4 (conditional expectation): Let X andY be two r.v.'s dened on the same probability space. Lesson 26 Linearity of Expectation | Introduction to Probability y & 0 & 1 \\ Therefore, the properties enjoyed by the expected value, such as linearity, are also enjoyed by the conditional expectation. PDF Lecture 2 : Conditional Expectation II ErX I As PrrAs if PrrAs0; 0 otherwise: 8/63. \], \[ E[X] = \underbrace{\frac{N_1}{N} + \frac{N_1}{N} + \ldots + \frac{N_1}{N}}_{\text{$n$ terms}} = n \frac{N_1}{N}. &= E\left(\sum_{i=1}^{k} a_iX_i+a_{k+1}X_{k+1} \middle| Y=y\right)\\ is "life is too short to count calories" grammatically wrong? JW+`V*Z,cWk($ W6D. It only takes a minute to sign up. Then, as shown in the proof of Theorem 2.2.5, for all x. r.p(x) = maxl(x) IEC For a random variable X and for all f E C, we have r.p(X) ~ f(X) and, This probability is \(E[Y_{ij}] = \frac{N_1^2}{N^2}\). In other words E[y i] = EfE[y ijX i]g; (3.1.1) where the outer expectation uses the distribution of X i. (This follows by the linearity of conditional expectation and the monotone convergence theo- rem, as you should check.) Independence concept. E[aX] &= aE[X] \tag{26.1} \\ which holds with strict inequality if $h(x) \neq E(Y \mid X)$. \hline Since there are \(n(n-1)\) \(Y_{ij}\)s, More- E[Z | N] &= E[Y_1 + \dots + Y_N| N] \\ If X X is a Binomial(n,N 1,N 0) Binomial ( n, N 1, N 0) random variable, then we can break X X down into the sum of simpler random variables: X = Y 1 +Y 2 ++Y n, X = Y 1 + Y 2 + + Y n, where Y i Y i represents the outcome of the i i th draw from the box. The conditional expectation as its name suggest is the population average conditional holding certain variables fixed. 0 What to throw money at when trying to level up your biking from an older, generic bicycle? Properties of conditional expectation Properties of conditional expectation (a) Linearity. 5.6 Conditional expected value. gzj1e)iPpq`T4A%TO G>%l // z(D:dZL*31MC ahcO|&a+;`cQ:`qC~!Tcn @gIHLD69. 6>EWZ'XuY n1dR{7Wf2A$yEZ% Hrm(uFdL}cbc WHH +8$z2Qpg["h>,CMoWkJ-.1YJ*w!r+(S`Z{'*,}'tLz{ E[aX] &= \sum_x ax f(x) & \text{(LOTUS)} \\ \[ E[Y - X] = E[Y] + E[-1 \cdot X] = E[Y] + (-1) E[X] = E[Y] - E[X]. It might help to work with just two joint random variables before you generalize. &+\underbrace{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_{k+1}x_{k+1})~f_{X_1,,X_k,X_{k+1}|Y}(x_1,,x_{k+1}|y)~dx_1dx_{k+1}\\ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. The above can be re-phrased verbatim for, say, $\mathbb{R}^n$: Let $Y \in \mathbb{R}^n$ and $V$ be a subspace. So in some sense it doesn't really matter whether $E$ is $E_{YX}$ or $E_{Y|X}$. \end{align*} $$ Why Does Braking to a Complete Stop Feel Exponentially Harder Than Slowing Down? So, $$\arg \min_{g(x)} E\left[Y-g(X)\right]^2 = \arg \min_{g(x)} \Big\{ -2g(X)E(Y\mid X) + \Big[g(X)\Big]^2 \Big\}$$. stream Linearity Proposition If a and b are real numbers and X and Y are random variables, then EraX bY |As aErX |As bErY |As: . Let and be integrable random variables, 0 and c, c1, c2 be real numbers. But I realize this is not just about the sign before the 2. We can think of \(W\) as an indicator variable for whether or not we win. &= \sum_x x f_X(x) + \sum_y y f_Y(y) & \text{(definition of marginal distribution)} \\ connecting the two tickets in the diagram above. Also note that the outer expectation is conditional on $X$. gm0)isCGOGt0=>%R51uDB&S#)Z[ f0 0u=ePV-AKs_ 8O~JB#2`V) x\[~_G:`AqNr7e@kNEB`07prgO9=HHcMf3BR:D!B|j$;vEXRbU+.5_9#`b}S7#h &= N\cdot E[Y] Doesn't make sense, as $g(X)$ is a random variable if $E$ is $E_{XY}$ and not $E_{Y|X}$. Show you should really write $E\Big[\big(Y - g(X)\big)^2|X\Big]$ or $E_{Y|X}\Big[\big(Y - g(X)\big)^2\Big]$ to make this clear. How is conditional expectation related to projection? corr(X,Y) = 1 Y = aX + b for some constants a and b. As I explained, my understanding of the proof leads me to blatantly problematic statement. The short answer is yes: one example is the Compound Poisson distribution. (Just as in the finite dimensional case, minimizing $L^2$-distance to a subspace means finding the projection). Asking for help, clarification, or responding to other answers. PDF Conditional expectation - uni-frankfurt.de (Hint: Follow Example 26.4. From Integral of Integrable Function is Additive, we have: X + Y is Pr -integrable. derivative of conditional expectation. We show how to think about a conditional expectation E(Y|X) of one r.v. Linearity of Expectation: Let R 1 and R 2 be two discrete random variables on some probability space, then. Then: Here is a using 2D LOTUS (25.1), with \(g(x, y) = x + y\). 2 0 obj << Theorem 1 (Expectation) Let X and Y be random variables with nite expectations. \[ \begin{array}{r|cc} w & 0 & 1 \\ \hline f_W(w) & 37/38 & 1/38 \end{array}. ), \[\begin{align} PDF Linearity of Expectation - IIT Delhi Their joint distribution &= a E[X] & \text{(definition of expected value)}. Example 26.2 (Xavier and Yolanda Revisited) Xavier and Yolanda head to the roulette table at a casino. \end{array}. Conditional mean in general is defined for $L^1$ random variables, which is a larger than $L^2$. &= E[X] + E[Y] & \text{(definition of expected value)}. with each persons phone number stored in a random location (independently), 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, $2 E\Big[ \big(Y - E(Y|X)\big) \big(E(Y|X) - g(X)\big)\Big] = 0$, $E \Big[ 2 \big(Y - E(Y|X)\big) \big(E(Y|X) - g(X)\big) + \big(E(Y|X) - g(X)\big)^2\Big]$, $E \Big[ 2 \big(Y - E(Y|X)\big) \big(E(Y|X) - g(X)\big) + \big(E(Y|X) - g(X)\big)^2\Big] = E( 0 + 0)$.

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