View Full Version : Covariance and Mean Absolute Deviation

feral

July 16th 2008, 05:39 PM

Hey. I have some questions and there are parts of the answers that I don't get....any help would be great!

1. Suppose X and Y are bivariate normal r.v. with EX=EY=1 and VARX=VARY=4. If Cov(X,Y) =3, find p(2X-3Y)<0.

So I know E(2X-3Y)= 2EX-3EY= -1

This is the part I'm confused about:

Var(2X-3Y)= Var(x) + Var(y) + 2Cov(X,Y)

= 2*2Var(x) -2*3*2Cov(X,Y) +9Var(y).

I'm confused as to where the 9Var(y) came from.

2. The mean absolute deviation of r.v. X is given by E abs(X-m(X)), where m(x) is the median of x. Find the mean absolute deviation of the poisson r.v. with mean 1.5.

I just need some assistance with this one. :skeptical:

NoMoreExams

July 16th 2008, 07:09 PM

Hey. I have some questions and there are parts of the answers that I don't get....any help would be great!

1. Suppose X and Y are bivariate normal r.v. with EX=EY=1 and VARX=VARY=4. If Cov(X,Y) =3, find p(2X-3Y)<0.

So I know E(2X-3Y)= 2EX-3EY= -1

This is the part I'm confused about:

Var(2X-3Y)= Var(x) + Var(y) + 2Cov(X,Y)

= 2*2Var(x) -2*3*2Cov(X,Y) +9Var(y).

I'm confused as to where the 9Var(y) came from.

You need to study your formulas. The bolded one is very inaccurate.

Learn this and prove to yourself that

V[aX + bY] = a^2*V(X) + b^2*V(Y) + 2*a*b*Cov(X,Y)

2. The mean absolute deviation of r.v. X is given by E abs(X-m(X)), where m(x) is the median of x. Find the mean absolute deviation of the poisson r.v. with mean 1.5.

I just need some assistance with this one. :skeptical:

First hint, what's the median of a Poisson R.V. with mean 1.5?

feral

July 16th 2008, 09:31 PM

Okay, thanks!

Very helpful...I don't know where I was getting that formula from. Whoops.

Also, one more question. This is simple, but I've done it over and over and keep getting the same answer:

A continuous random variable has density function

f(x) = (1/9) (3-abs(x)) for abs(x)<3 and 0 otherwise

Find the Variance.

I keep getting 6 for EX and -2 for EXsquared...The answer is supposed to be (3/2)... help

NoMoreExams

July 17th 2008, 01:37 AM

Okay, thanks!

Very helpful...I don't know where I was getting that formula from. Whoops.

Also, one more question. This is simple, but I've done it over and over and keep getting the same answer:

A continuous random variable has density function

f(x) = (1/9) (3-abs(x)) for abs(x)<3 and 0 otherwise

Find the Variance.

I keep getting 6 for EX and -2 for EXsquared...The answer is supposed to be (3/2)... help

f(x) = (3 - |x|)/9 for -3 < x < 3

so we can rewrite it as

f(x) = (3 - x)/9 for [0,3]

(3 + x)/9 for [-3,0]

so E(X) = 1/9 * [integral(3x - x^2 dx from 0 to 3) + integral(3x + x^2 dx from -3 to 0)]

= 1/9 * [(3x^2/2 - x^3/3 evaluated between 0 and 3) + (3x^2/2 + x^3/3 evaluated beetween -3 and 0)]

= 1/9 * [(27/2 - 9) + (0 - 27/2 + 9)] = 0

Similarly

E(X^2) = 1/9 * [integral(3x^2 - x^3 from 0 to 3) + integral(3x^2 + x^3 from -3 to 0)]

= 1/9 * [(x^3 - x^4/4 evaluated between 0 and 3) + (x^3 + x^4/4 evaluated between -3 and 0)]

=1/9 * [(27 - 81/4) + (0 + 27 - 81/4)] = 1/9 * [54 - 40.5] = 13.5/9 = 3/2

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