urgent hw....do help....probability densities

• Nov 9th 2007, 03:08 AM
souhar
urgent hw....do help....probability densities
Hi i have the following problems as HW.......try to help me......urgent.......

1) If X and Y have a bivariate normal distribution and U=X+Y and V=X-Y find an expression for the correlation coefficient of U and V.
2)If X has an exponential distribution,show that
P(X >= t + T \ X >= T) = P(X >= t)
this property of an exponential random variable parallels that of a geometric random variable as [P(X = x+n \ X > n)] = P (X=x)
3)If rando variable T is the time to failure of a commercial product and the values of its probability dnsity and distribution function at time "t" are f(t) and F(t), then its failure rate at time t is given by f(t) / 1-F(t). Thus the failure rate at time t is the probability density of failure at time t given that failure does not occur prior to time t.
a) show that if T has an exponential distribution, the failure rate is constant.
b)show that if has a weibull distribution, the failure rate is given by ab t^b-1.

• Nov 9th 2007, 10:59 AM
CaptainBlack
Quote:

Originally Posted by souhar
1) If X and Y have a bivariate normal distribution and U=X+Y and V=X-Y find an expression for the correlation coefficient of U and V.

$\displaystyle \rho_{U,V} = \frac{E((U-\bar{u})(V-\bar{v}))}{\sigma_U\sigma_V}= $$\displaystyle \frac{E((X+Y-\bar{x}+\bar{y})(X-Y-\bar{x}+\bar{y}))}{\sigma_U\sigma_V}=$$\displaystyle \frac{E((X-\bar{x})^2+(Y-\bar{y})^2)}{\sigma_U\sigma_V}= $$\displaystyle \frac{\sigma_X^2+\sigma_Y^2}{\sigma_U\sigma_V} Now: \displaystyle \sigma^2_U = E((U-\bar{u})^2)=$$\displaystyle E( [(X+Y) - (\bar{x}+\bar{y})]^2)=$$\displaystyle E((X-\bar{x})^2 + 2(X-\bar{x})(Y-\bar{y}) +(Y-\bar{y})^2)=$$\displaystyle \sigma_X^2 + 2\sigma_{X,Y} + \sigma_Y^2$

A similar argument gives:

$\displaystyle \sigma^2_U = \sigma_X^2 - 2\sigma_{X,Y} + \sigma_Y^2$

and I leave the rest to you.

(By the way I have not used the information that X and Y are bivariate normal)

RonL