The conditional expectation E ( X | Y = y) is shorthand for E ( X | { Y = y }). 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. Manage Settings We are often interested in the expected value of a sum of random variables. For example, suppose we are . 9.11. 4.29K subscribers PROBABILITY & STATISTICS PLAYLIST: https://goo.gl/2z3jX6 _____________ In this video you will learn how to use the Geometric Distribution and Conditional Probability to find the. \(P(X = i) = 1/n\), \(E[Y|X = i] = i/6\), so, \(E[Y] = \dfrac{1}{6} \sum_{i = 0}^{n} i/n = \dfrac{(n + 1)}{12}\), b. n = 100; p = 1/6; X = 1:n; Y = 0:n; PX = (1/n)*ones(1,n); P0 = zeros(n,n+1); % Could use randbern for i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i); end P = rot90(P0); jcalc EY = dot(Y,PY) EY = 8.4167 % Comparison with part (a): 101/12 = 8.4167. This is already given in the total column of our table: 0.03 + 0.15 + 0.15 + 0.16 = 0.49. As it's a random variable, it has an expectation, which we can calculate using the non-linear function rule. Depending on the context, the conditional expectation can be either a random variable or a function. Conditional Expectation, Regression, Applied Probability 2009 - Paul E Pfeiffer | All the textbook answers and step-by-step explanations Samy T. Conditional expectation Probability Theory 2 / 64 Outline. The classical gambler's ruin problem was solved by Abraham de Moivre over two hun-dred years ago, using a method that has grown into one of the main technical tools of mod-ern probability. \(P(X = i) = 1/(n + 1)\), \(0 \le i \le n\), \(P(Y = k|X = i) = 1/(i + 1)\). Take two discrete variables and and consider them jointly as a random vector Suppose that the support of this vector is and that its joint pmf is Let us compute the conditional pmf of given . This example demonstrates the method of conditional probabilities using a conditional expectation. b. Put \(p = \mu/(\mu + \lambda)\) and \(q = 1 - p = \lambda/(\mu + \lambda)\) to get the desired result. (mz1'CXAOXS_Cv/qZ}\:. probability probability-theory expectation. Independence concept. Use the fact that \(g(X, Y) = g^* (X, Y, Z)\), where \(g^* (t, u, v)\) does not vary with \(v\). A.2 Conditional expectation as a Random Variable Conditional expectations such as E[XjY = 2] or E[XjY = 5] are numbers. We prefer to think about ErX |Ysas a function R. Determine the joint distribution for \(\{X, Y\}\) and then determine \(E[Y]\). \(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \text{min } \{1, 2 - t\}\) (see Exercise 38 from "Problems on Mathematical Expectaton", Exercise 14.2.8). Of course, most analysts are more comfortable with functions rather than measures and prefer to use the Radon-Nikodym theorem with respect to the pushforward of to convert the pushforward of f to an integrable function. Use jcalc to determine \(E[Y]\); compare with the theoretical value. \(P(X = i, Y = k) = 1/(n + 1)(i + 1)\), \(0 \le i \le n\), \(0 \le k \le i\). &oXG This implies that X + Y Gamma(2,). Then taking we obtain (10.8), and hence, the necessity property follows. The conditional expectation of a positive random variable is positive. To find the distribution of X+Y we have to find the probability of the sum by using the conditioning as follows. What is the probability the failure is due to the \(i\)th component? The function form is either denoted or a separate function symbol such as is introduced with the meaning . First-step analysis for calculating eventual probabilities in a stochastic process. Determine \(E[Y]\) from \(E[Y|X = i]\). 1,186. Same as Exercise 14.2.20, except \(p = 1/10\). The expectation of a random variable is the long-term average of the . So it is a function of y. Legal. Lecture 10: Conditional Expectation 10-2 Exercise 10.2 Show that the discrete formula satis es condition 2 of De nition 10.1. Let's connect through LinkedIn - https://www.linkedin.com/in/dr-mohammed-mazhar-ul-haque-58747899/, Is Lonely Adjective Or Noun? Some of our partners may process your data as a part of their legitimate business interest without asking for consent. 2X#9R~X/!B]^xs$(D{l|JbYA!]TDai$h)Tv5C}Utj}oQ0h0>k Data were kept on the effect of training time on the time to perform a job on a production line. Find Var ( Z). endobj Posted on February 13, 2014 by Jonathan Mattingly | Comments Off. /Length 3543 \dfrac{\mu^k \lambda^{n - k}}{(\mu + \lambda)^n}\). We add and subtract two terms, E[Y X], do a little algebra, and show that the cross term goes to zero: E[(Y g(X))2] = E[(Y E[Y X])2]+E[(E[Y X] g(X))2] +2E[(Y E[Y X])(E[Y X]g(X))] (2) Using two properties, 7.1. Limit Theorems 307. Suppose the parameter is the value of random variable \(H\) ~ uniform on[0.005, 0.01], and \(X\) is conditionally exponential \((u)\), given \(H = u\). Thus, in this example the conditional expectation is the measure on (X,F,) that assigns the number 2 (the sum of 1/2^x) to X. Exercises 283. These random variables can be represented by a random vector X that . |8n7CdSWf=[?fR<=g(Zy6/
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CPN#Drg a conditional expectation is an immediate consequence of the Radon-Nikodym theorem. The conditional expectation works out to 4*1/3 + 5*1/3 + 6*1/3 = 5 In a similar manner, once we determine the loss amounts for each data point, we factor in the condition that the loss amount exceeds 68.77, i.e. Use of linearity, (CE8), and (CE10) gives, \(E[Z|X = t] = I_M (t) 4t + I_N(t) (t + E[Y|X = t])\), \(= I_M (t) 4t + I_N (t) (t + \dfrac{(t + 1)(t + 3) (3t + 1)}{4(1 + 4t + t^2)})\). This article will help to find whether lonely is an adjective or a noun. Massimo Banzi with Arduino. (See Exercise 17 from " Problems On Random Vectors and Joint Distributions", Exercise 27 from "Problems on Mathematical Expectation", and Exercise 27 from "Problems on Variance, Covariance, Linear Regression"). Solution: Step 1: Find the sum of the "given" value (X = 1). \(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2t, 3 - t\}\). stream We can use the word "lucky" to mark good fortune most of the time. 1.4.5 Solved Problems: Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. the VaR loss amount. Suggestion. (See Exercise 17 from "Problems on Mathematical Expectation"). xg{ ;OIdLg47*k#Z 3s{ Zz|A @nM{q^/-F~ >vqe5p26RX$oq9.@endstream The regression line of \(Y\) on \(X\) is \(u = (4t + 5)/9\). There are 60 Suv's, 40 Sport cars, 50 Vans and 50 Coupe, a total of 200 cards. A shop which works past closing time to complete jobs on hand tends to speed up service on any job received during the last hour before closing. /Filter /FlateDecode The bottom line will be that, in many important respects, conditional expectations behave like ordinary expectations, with random quantities that are functions of the conditioning random variable being treated as constants.2 Let Y be a random variable, vector, or object valued in a measurable space, and let . 684 The derivation is nearly the same as above. \(E[Y|X = t] = \dfrac{2}{3} (1 - t)\) for \(0 \le t < 1\) and \(X\) has density function \(f_X (t) = 30 t^2 ( 1 - t)^2\) for \(0 \le t \le 1\). a. \(E[Y] = \int E[Y|X =t] f_X (t)\ dt = \int_{0}^{1} 20t^2 (1 - t)^3\ dt = 1/3\). \(F_{X|H} (t|u) = 1 - e^{ut}\) \(f_{H} (u) = \dfrac{1}{0.05} = 200\), \(0.005 \le u \le 0.01\), \(F_X (t) = 1 - 200 \int_{0.005}^{0.01} e^{-ut}\ du = 1 - \dfrac{200}{t} [e^{-0.005t} - e^{-0.01t}]\), \(P(X > 150) = \dfrac{200}{150}[e^{-0.75} - e^{-1.5}] \approx 0.3323\), \(E[X|H = u] = 1/u\) \(E[X] = 200 \int_{0.005}^{0.01} \dfrac{du}{u} = 200 \text{ln } 2\), A system has \(n\) components. 2 Conditional expectation with respect to a random variable. As usual, let 1A denote the indicator random variable of A. Suppose that (W,F,P) is a probability space where W = fa,b,c,d,e, fg, F= 2W and P is uniform. Answer Exercise 14.2.22 E[Y | X = t] = 10t and X has density function fX(t) = 4 2t for 1 t 2. \(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2\), for \(0 \le u \le 1\). The conditional Expectation for the discrete and continuous random variable with different examples considering some of the types of these random variables discussed using the independent random variable and the joint distribution in different conditions, Also the expectation and probability how to find using conditional expectation is explained with examples, if you require further reading go through below books or for more Article on Probability, please follow our Mathematics pages. Out of the framework of Linear Theory, a signicant role plays the independence concept and conditional expectation. 1jh|j*$0L#,srgA}O;8yS#(Vu$G Let X be the random variable that denote the time in hours until the person came out safely and Y denote the pipe he chooses initially, so, If the person chooses the second pipe , he spends 5 hous in that but he come outside with expected time, Let N be the random number of random variable and sum of random variables is then the expectation, If the probability density function of the bivariate random variable X and Y is, then the correlation between random variable X and Y for the bivariate distribution with density function is, since the expectation using conditional expectation is, for the normal distribution the conditional distribution X given Y is having mean, In the geometric distribution let us perform successively independent trials which results in success with probability p , If N represents the time of first success in these succession then the variance of N as by definition will be, Let the random variable Y=1 if the first trial results in success and Y=0 if first trial results in failure, now to find the mathematical expectation here we apply the conditional expectation as, if success is in first trial then N=1 and N2=1 if failure occur in first trial , then to get the first success the total number of trials will have the same distribution as 1 i.e the first trial that results in failure with plus the necessary number of additional trials, that is, since the expectation of geometric distribution isso, so the variance of geometric distribution will be, The sequence of uniform random variables U1, U2 .. over the interval (0, 1) and N is defined as, then for the expectation of N, for any x [0, 1]the value of N, to find the expectation we use the definition of conditional expectation on continuous random variable, now conditioning for the first term of the sequence we have. Determine the joint distribution and then determine \(E[Y]\). Example: Suppose = [0;1], C is the Cantor set and f:[0;1]! It is also known as conditional expected value or conditional mean . Imagine that you randomly select one of these denominational values from $5 to $100 (i.e., $10) and withdraw it from your bank. e^{-\lambda} \dfrac{\lambda^{n -k}}{(n - k)! We and our partners use data for Personalised ads and content, ad and content measurement, audience insights and product development. 7. The regression line of \(Y\) on \(X\) is \(u = -0.1359 t + 1.0839\). \(f_X (t) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\), \(f_{Y|X} (t|u) = I_{[0, 1]} (t) \dfrac{t^2 + 2u}{t^2 + 1} + I_{(1, 2]} (t) 3u^2\) \(0 \le u \le 1\), \(E[Y|X = t] = I_{[0, 1]} (t) \dfrac{1}{t^2 + 1} \int_{0}^{1} (t^2u + 2u^2)\ du + I_{(1, 2]} (t) \int_{0}^{1} 3u^3 \ du\), \(= I_{[0, 1]} (t) \dfrac{3t^2 + 4}{6(t^2 + 1)} + I_{(1, 2]} (t) \dfrac{3}{4}\), For the distributions in Exercises 12-16 below. Suppose the arrival time of a job in hours before closing time is a random variable \(T\) ~ uniform [0, 1]. Find the Marginal PMFs of X and Y. To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. \(f_X(t) = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23} t^2\), \(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\), \(I_M (t, u) = I_{[0, 1]} (t) I_{[0, 1]} (u)\) \(I_{M^c} (t, u) = I_{[0, 1]} (t) I_{[1, 2 -t]} (u) + I_{(1,2]} (t) I_{[0, 1]} (u)\), \(E[Z|X = t] = I_{[0, 1]} (t) \dfrac{1}{2(2 - t)} \int_{0}^{1} (t + u) (t + 2u)\ du + \dfrac{1}{2 - t} \int_{1}^{2 - t} u (t + 2u)\ du] + I_{(1, 2]} (t) 2E [Y|X = t]\), \(= I_{[0, 1]} (t) \dfrac{1}{12} \cdot \dfrac{2t^3 - 30t^2 + 69t - 60}{t - 2} + I_{(1, 2]} (t) \dfrac{7}{6} 2t\). Conditional Expectation. \(F_Y (v) = \int F_{Y|T} (v|u) f_T (u)\ du = \int_{0}^{1} (1 - e^{-\beta (2 - u)v})\ du = \), \(1 - e^{-2\beta v} \dfrac{e^{\beta v} - 1}{\beta v} = 1 - e^{\beta v} [\dfrac{1 - e^{-\beta v}}{\beta v}]\), \(0 < v\). @ (ayUEv?3=;%Zvvz_78hl
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*K{[Vw,,$!.m (9(|Vj2}d ]-G+ mX" https://en.wikipedia.org/wiki/probability_distribution, A first course in probability by Sheldon Ross, Schaums Outlines of Probability and Statistics, An introduction to probability and statistics by ROHATGI and SALEH, I am DR. Mohammed Mazhar Ul Haque. That is to . The regression line of \(Y\) on \(X\) is \(u = 0.5275t + 0.6924\). % The object is to nd expressions for and that are in terms of the rst-order and second-order moments of the joint distribution. The regression line of \(Y\) on \(X\) is \(u = -t/11 + 35/33\). Conditional Expectation Problem. The regression line of \(Y\) on \(X\) is \(u = (-124t + 368)/431\), \(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{11} t + I_{(1, 2]} (t) \dfrac{12}{11} t (2 - t)^2\), \(f_{Y|X} (u|t) = I_{[0, 1]} (t) 2u + I_{(1, 2]} (t) \dfrac{2u}{(2 - t)^2}\), \(E[Y|X = t] = I_{[0, 1]} (t) \int_{0}^{1} 2u^2 \ du + I_{(1, 2]} (t) \dfrac{1}{(2 - t)^2} \int_{0}^{2 - t} 2u^2 \ du\), \(= I_{[0, 1]} (t) \dfrac{2}{3} + I_{(1, 2]} (t) \dfrac{2}{3} (2 - t)\). \(f_{X} (t) = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]} (t) 6t^2 (1 - t^2)\). Having the immense ability of problem design and solving. Let \(Y\) be the number of sevens thrown on the \(X\) tosses. >> If you would like to change your settings or withdraw consent at any time, the link to do so is in our privacy policy accessible from our home page. \(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\), \(f_{Y|X} (t|u) = I_{[0, 1]} (t) \dfrac{t + 2u}{6t^2} + I_{(1,2]} (t) \dfrac{t + 2u}{3(3 - t)}\) \(0 \le u \le \text{max } (2t, 3 - t)\), \(E[Y|X = t] = I_{[0, 1]} (t) \dfrac{1}{6t^2} \int_{0}^{t} (tu + 2u^2)\ du + I_{(1, 2]} (t) \dfrac{1}{3(3 - t)} \int_{0}^{3 - t} (tu + 2u^2)\ du\), \(= I_{[0, 1]} (t) \dfrac{11}{9} t + I_{(1, 2]} (t) \dfrac{1}{18} (t^2 - 15t + 36)\). \(E[Y|X = t] = 10t\) and \(X\) has density function \(f_X (t) = 4 - 2t\) for \(1 \le t \le 2\). 6 Conditional regular laws. In probability theory, a conditional expectation is the expected value of a real random variable with respect to a conditional probability distribution. a. 6.1 Conditional Distribution 261. Time to failure \(X\) of a manufactured unit has an exponential distribution. \(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\), \(I_M(t, u) = I_{[0, 1]} (t0 I_{[1, 2]} (u)\) \(I_{M^c} (t, u) = I_{[0, 1]} (t) I_{[0, 1)} (u) + I_{(1, 2]} (t) I_{[0, 3 - t]} (u)\), \(E[Z|X = t] = I_{[0, 1/2]} (t) \dfrac{1}{6t^2} \int_{0}^{2t} 2u^2 (t + 2u) \ du +\), \(I_{(1/2, 1]} (t) [\dfrac{1}{6t^2} \int_{0}^{1} 2u^2 (t + 2u)\ du + \dfrac{1}{6t^2} \int_{1}^{2t} (t + u) (t + 2u)\ du] + I_{(1, 2]} (t) \dfrac{1}{3 (3 - t)} \int_{0}^{3 - t} 2u^2 (t + 2u)\ du\), \(= I_{[0, 1/2]} (t) \dfrac{32}{9} t^2 + I_{(1/2, 1]} (t) \dfrac{1}{36} \cdot \dfrac{80t^3 - 6t^2 - 5t + 2}{t^2} + I_{(1, 2]} (t) \dfrac{1}{9} (- t^3 + 15t^2 - 63t + 81)\). Lecture 10: Conditional Expectation 2 of 17 Example 10.2. Capable of Motivating candidates to enhance their performance. By using conditional expectation and conditional probability. 9 0 obj Multivariate Normal Distribution 291. As the next example illustrates we sometimes use the conditional distribution of a random variable to compute the (unconditioned) distribution of the random variable. This is a somewhat subtle difference. The Industrial Talk Podcast With Scott MacKenzieMarc O'Regan, CTO Dell Technologies199! Jon Clay with Trend Micro. Conditional Expectation As a Projection. \(f_X (t) = 2(1 - t)\), \(0 \le t \le 1\), \(f_{Y|X} (u|t) = \dfrac{1}{2(1 - t)}\). Therefore, E(X) = {E(X|Si)P (Si)} E ( X) = { E ( X | S i) P ( S i) } Example The probability of relaxed trade restrictions in a given country is 40%. The consent submitted will only be used for data processing originating from this website. Determine \(P(X > 150)\). Determine \(E[Y]\). In this chapter, we look at the same themes for expectation and variance. \(X\) is the amount of training, in hours, and \(Y\) is the time to perform the task, in minutes. Lawrence Leemis. * )#)!H5$xoW;#4A} %!cg[i z70;"SxZ6Q5 > 64h:o)J"h%Y`{0i$" 3 Facts(When & Examples). Consider the following probability space where . Example 2. 2. Determine \(E[Y]\) from \(E[Y|X = k]\). 4 Conditional expectation: properties. Example Let the support of the random vector be and its joint probability mass function be Let us compute the conditional probability mass function of given . A number \(X\) is selected randomly from the integers 1 through 100. 3. Conditional probability: P(A|B). The random variable is denoted analogously to conditional probability. In similar way if X and Y are continuous then the conditional probability density function of the random variable X given Y is, where f(x,y)is joint probability density function and for all yfY(y)>0 , so the conditional expectation for the random variable X given y will be, As we know that all the properties of probability are applicable to conditional probability same is the case for the conditional expectation, all the properties of mathematical expectation are satisfied by conditional expectation, for example conditional expectation of function of random variable will be, and the sum of random variables in conditional expectation will be, To find conditional expectation of the sum of binomial random variables X and Y with parameters n and p which are independent, we know that X+Y will be also binomial random variable with the parameters 2n and p, so for random variable X given X+Y=m the conditional expectation will be obtained by calculating the probability, thus the conditional expectation of X given X+Y=m is, if the joint probability density function of continuous random variables X and Y is given as, To calculate the conditional expectation we require conditional probability density function, so, since for the continuous random variable the conditional expectation is, hence for the given density function the conditional expectation would be, Expectation by conditioning||Expectation by conditional expectation, We can calculate the mathematical expectation with the help of conditional expectation of X given Y as, for the discrete random variables this will be, and for the continuous random we can similarly show. Our choice generalizes nicely. Each of two people draw \(X\) times, independently and randomly, a number from 1 to 10. The conditional probability of an event A, given random variable X (as above), can be defined as a special case of the conditional expected value. STA 205 Conditional Expectation R L Wolpert a(dx) = Y(x)dx with pdf Y and a singular part s(dx) (the sum of the singular-continuous and discrete components). By (CE9), \(E[g(X, Y)|Z] = E\{E|g(X, Y)|X, Z]|Z\} = E[e(X, Z)|Z]\) a.s. \(E[e(X, Z)|Z = v] = E[e(X, v)|Z = v] = \), \(\int E[g(X, Y)|X = t, Z = v] F_{X|Z} (dt|v) = \), \(\int E[g(t, Y)|X = t, Z = v] F_{X|Z} (dt|v)\) a.s. \([P_Z]\). 6.2 Conditional Expectation given a -algebra 276. 1,078 Apart from two minor points, your solution is correct: . The word "lucky" comes We are group of industry professionals from various educational domain expertise ie Science, Engineering, English literature building one stop knowledge based educational solution. In many problems, we are interested in more than one random . We answer the questions on finding conditional probabilities using two methods: 1) the definition and 2) restriction of the sample space. 6. the remaining number of uniform random variable is same at the point where the first uniform value is y,in starting and then were going to add uniform random variables until their sum surpassed x y. so using this value of expectation the value of integral will be, and m(1) =e, the expected number of uniform random variables over the interval (0, 1) that need to be added until their sum surpasses 1, is equal to e, We can find the probability also by using conditional expectation like expectation we found with conditional expectation, to get this consider an event and a random variable X as, from the definition of this random variable and expectation clearly, now by conditional expectation in any sense we have. b. So one of the things we will do here is redo the L 2 version so as to give us an alternative proof of the existence of conditional expectation. Here's the problem : Consider the probability space = [0, 1] with the $\. with continuous distribution (and finite mean). Example 1: Weather Forecasting One of the most common real life examples of using conditional probability is weather forecasting. Let X, Y and Z be . (n - k)!} The regression line of \(Y\) on \(X\) is \(u = 1.0561 t - 0.2603\). n = 50; X = 1:n; Y = 1:n; P0 = zeros(n,n); for i = 1:n P0(i,1:i) = (1/(n*i))*ones(1,i); end P = rot90(P0); jcalc: P X Y - - - - - - - - - - - - EY = dot(Y,PY) EY = 13.2500 % Comparison with part (a): 53/4 = 13.25. 8. Determine the distribution unction for \(Y\). Example: Roll a die until we get a 6. \(0 \le t \le 1\), \(0 \le u \le 2(1 - t)\), \(E[Y|X = t] = \dfrac{1}{2(1 - t)} \int_{0}^{2(1-t)} udu = 1 - t\), \(0 \le t \le 1\). xXK5=v* We can also compute conditional expectations where we condition on another random variable not the event that this other random variable takes on a value. Let \(Y\) be the number of matches (i.e., both draw ones, both draw twos, etc.). { "14.01:_Conditional_Expectation_Regression" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.
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