Recent content by ja404

I figured out one thing I've been doing wrong. My new answer for pdf of Y=min(X1,X2) is 4e^-4y I'm pretty confident about that one.

Thank you, that's all I need. Mods can close.

[SOLVED] Properties of variance Would the Var(aX + bY + c) just be the Var(aX+bY) since adding a single number to the function doesn't change the variance. I would then be able to use the property: Var(aX+bY)= a^2Var(X)+b^2Var(Y)+2abCov(X,Y) Just wondering if anyone can confirm my...

Any ideas on what I might be doing wrong? Thanks.

well after talking to my professor, he said I should be using the order statistic technique pdf of Y=min(X1,X2...Xn) = g(y) = n*[(1-Fx(y))^n-1]*fx(y) where n=2 for this case Fx(y) is cdf of X evaluated at y = -e^-2y fx(y) is pdf of X evaluated at y = 2e^-2y I put those into g(y) and...

Ok, I'm coming up with a pdf of: (4e^-2y)*(1+e^-2y) OR (4e^-2y)+(4e^-4y) Thoughts?

Would it correct to say then P(Y1>y) = P(X1>Y)*P(X2>Y) ? Thanks.

I just can't figure out how to deal with that "min". I know that if say Y=X1^2 then I would just follow: cdf = Fy(y) = P(Y<y) = P(X^2<y) = P(X<Y^1/2) etc, but I don't know how to represent that 'minimum' when I'm trying to solve the inequality for X. Thanks again.

Thanks for the help. I don't recall covering the dirac delta at all but I'm trying to use it and we'll see how it works. I was thinking there might be a transform involved as well? Thanks again, I appreciate it.

Homework Statement The lifetime of batteries are independent random variables Xi, i= 1,2,... each having exponential distribution given by the density f(x) = 2e^-2x, x>0, 0 elsewhere. If a flashlight needs two batteries to work, then the time that the flashlight can operate is a random...