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New York University

If X and Y are independent find E[ XY ]



Random Variability: Covariance and Correlation

%20covariance%20and%20correlation.pdf



Chapter 3: Expectation and Variance

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28 Feb 2011 Let x and y be independent random variables with means µx and µy and variances ?2 x and ?2 y respectively. Show that. Var[xy] = ?2.



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Notice that the variance of X is just the covariance E(XY ) ? µXE(Y ) ? E(X)µY + µXµY ... If X and Y are independent variables then their.



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19 Sept 2005 This isn't the only time that E [XY ] = E [X] E [Y ] though. Here's where independence gets important: what's the variance of X + Y ?



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Chapter 5. Multiple Random Variables 5.4: Covariance and

helps us finally compute the variance of a sum of dependent random actually proved in 5.1 already that E [XY ] = E [X] E [Y ] when X Y are independent.



Practice problems — Solutions - University of Illinois

Since J K L are independent the moment-generating function for their sum X is equal to the product of the individual moment-generating functions i e M X(t) = M K(t)M J(t)M L(t) = (1?2t)?3?2 5?4 5 = (1?2t)?10 Di?erentiating this function we get M0(t) = (?2)(?10)(1?2t)?11 M00(t) = (?2)2(?10)(?11)(1?2t)?12

Is var(x+y)=var(X)+var (Y) when X and Y are two independent?

Let’s define X and Y to each be the result of the coin flip where we assign the value 1 to Heads and 0 to Tails. So X+Y take on the values 0,1 and 2 with probabilities 1/4,1/2 and 1/4. So So it is indeed true that Var (X+Y)=Var (X)+Var (Y) when X and Y are two independent random variables.

How to calculate var( x y) if X and Y are independent?

How i can calculate Var ( X Y) if X and Y are independent? I know this: Var ( X Y) = E ( X) 2 Var ( Y) + E ( Y) 2 Var ( X) + Var ( X) Var ( Y). But i need prove it.

Is there a block exogeneity in the VAR model?

There's no block exogeneity in your VAR model. Checking the p-values, I would suggest that in the first and second tests, the variables DRLM2 and DlRGDP have granger causality issues (since we reject for a 5% sig. level). In the third case, DRLM2 is not rejected for a sig. level of 5%, yet, it seems to be very close (4,8%).

What is experience vardaxyn and how does it work?

Vardaxyn is a revolutionary male enhancement product in the healthcare sector and it is essentially a new one designed to promote the health of your male body and sexual functions in a natural way. Vardaxyn is a natural ED cure supplement that heals all of the ejaculation problems in a completely natural way.

Inference and Regression

Assignment 3

Professor William Greene

Phone: 212.998.0876

Office: KMC 7-90

Home page:

www.stern.nyu.edu/~wgreene

Email:

wgreene@stern.nyu.edu

Course web page:

1. The random variable X has mean

x and standard deviation x . The random variable Y has mean y and standard deviation y . If X and Y are independent, find E[ XY ], Var[ XY ] and the standard deviation of XY.

If they are independent, E[XY] = E[X]E[Y] = x

y

Var[XY] = E[(XY)

2 ] - {E[XY]} 2 = E[X 2 Y 2 ] - (E[X]E[Y]) 2 = E[X 2 ]E[Y 2 ] - (E[X]) 2 (E[Y]) 2 x2 x2 y2 + y2 x2 y2 x2 y2 x2 y2 x2 y2 y2 x2 x2 y2 x2 y2 + x2 y2 y2 x2 The standard deviation is the square root of this.

2. The number of auto vandalism claims reported per month at Sunny Daze Insurance Company (SDIC)

has mean 100 and standard deviation 20. The individual losses have mean $1,200 /claim and standard deviation $80 /claim. The number of claims and the amounts of individual losses are independent.

Using the normal approximation, calculate the probability that SDIC's aggregate auto vandalism losses

reported for a month will be less than $110,000.

Aggregate losses are claims times loss per claim.

Expected value is the product of the means, 100claims $1,200/claim = $120,000

Variance based on question 1 is 202

80
2 + 20 2 1,200 2 + 80 2 100
2 = 642,560,000

Standard deviation = 25,349.

Prob[ losses < 110,000] = Prob[z < (110,000 - 120,000)/25,349] = Prob[z < -.394= ] =.347.

3. Let xbe the average of a sample of 26 independent normal random variables with mean 0 and variance

1. Determine c such that Prob(|

x| < c) = .5. Prob(|x-bar| < c) implies that the probability that x-bar is less than -c is .25 (and the probability that is it is greater than c is .25). X-bar is normally distributed with mean 1/sqr(16) = ¼ = .25. So,

Prob(x-bar < -c) = Prob[(xbar - 0)/.25 < -c/.25) = .25. From the normal table, -c/.25 = -.675. So,

4c = .675 or c = .16875.

Department of IOMS

4. Show that if X ~ F

n,m , then Y = 1/X ~ F m,n It is possible to do this problem by brute force, using a change of variable and the density of F n,m But, the result follows trivially from the definition of F. F n,m = [chi-squared(n) / n] / [chi-squared(m) / m].

Then, 1/F

n,m = [chi-squared(m) / m] / [chi-squared(n) / n]

5. Show that if T ~ t

n , then T 2 ~ F 1,n T = t n = Normal(0,1) / sqr[chi-squared(n)/n] T 2 = [Normal(0,1)] 2 / chi-squared(n)/n. The key now is to note that the square of a standard normal is chi-squared with 1 degree of freedom. The definition of F 1,n then follows immediately.

6. Suppose that X

1 ,X 2 ,...,X n are i.i.d. random variables on the interval [0,1] with density 1 21 (3 )( | )(1 ) , 0, 0 x 1.( ) (2 )fxx x The parameter is to be estimated from the sample. It can be shown that E[X] = 1/3 and

Var[X]=2/[9(3+1)].

a. How could the method of moments be used to estimate ? b. What equation does the MLE of satisfy. (I.e., what is the likelihood equation?) c. What is the asymptotic variance of the MLE of ? d. Find a sufficient statistic for a. You could not use the mean in the method of moments since E[X] does not depend on . So, use the variance. Equate s 2 to 2/[9(3+1)]. Solving, 2/s 2 = 9(3+1) or (2/9)/s 2 = 3 + 1, or [2/(9s 2 ) - 1]/3 = b. The log likelihood would be the sum of the logs of the density, logL = nlog(3) - nlog() - nlog(2) + (-1) i logx i + (2-1) i log(1-x i logL/ = 3n(3) - n() - 2n(2) + i logx i - 2 i log(1-x i ) = 0. c. (Ouch!). Differentiate it again. 2 logL/ 2 = 9n(3) - n() - 4n(2) = H The asymptotic variance is -1/H. Since H does not involve x i , we don't need to take the expected value.

d. There are two sufficient statistics that are jointly sufficient for . The density is an exponential

family, so we can see immediately that the sufficient statistics are i logx i and i log(1-x i The reason there are two statistics and one parameter is that this is a version of the beta distribution which generally has two parameters, and and the two sufficient statistics shown.

This distribution forces to equal 2.

7. Suppose that X

1 ,X 2 ,...,X n are an i.i.d. random sample from a Rayleigh distribution with parameter > 0, f(x|) = 22
2 exp[ /(2 )], x 0.xx a. Find a method of moments estimator of . b. Find the MLE of c. Find the asymptotic variance of the MLE of 2

222 22

2200
a. First find the expected value:quotesdbs_dbs4.pdfusesText_7
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