Recent content by cookiesyum

But I don't think the problem is saying X1 < 2X2 I think the phrasing was X2|X1 < 2X2. Am I thinking about it wrong?

Got it! Thanks so much for the help.

Yep, finally got the answer. Thank you everybody for your help!

Oh! I was doing it with three different t's ... t1 t2 and t3 which made it a million times harder. That makes sense. Thanks so much!

Homework Statement Let X1 X2 X3 be independent and identically distributed random variables with common pdf f(x) = e^-x 0<x<infinity, zero elsewhere. Evaluate P(X1 < X2|X1 < 2X2) and P(X1 < X2 < X3|X3 < 1). The Attempt at a Solution Would appreciate any hints as to where to start...

Homework Statement Let X1 X2 X3 and X4 be four independent random variables, each with pdf f(x) = 3(1-x)2, 0<x<1, zero elsewhere. If Y is the minimum of these four variables, find the cdf and pdf of Y. The Attempt at a Solution P(Y<or= y) = 1 - P(Y>y) = 1 - P(X1>y, X2>y, X3>y...

Homework Statement Let f(x1, x2, x3) = e-(x1+x2+x3), 0<x1,2,3<infinity, zero elsewhere be a joint pdf of X1, X2, X3. Compute P(X1< X2< X3) and P(X1= X2< X3) Determine the joint mgf. The Attempt at a Solution P(X1< X2< X3) = triple integral of e-(x1+x2+x3) dx2dx1dx3 as x2 goes from x1 to...

Homework Statement If D is open, and if f is continuous, bounded, and obeys f(p)>or=0 for all p in D, then the double integral over D of f is equal to 0 implies f(p) = 0 for all p. Homework Equations Hint: There is a neighborhood where f(p)>or=d. The Attempt at a Solution The...

Homework Statement Let X have a geometric distribution. Show that P(x>or= k+j|x>or=k) = P(X>or=j) where k and j are nonnegative integers. The Attempt at a Solution P(x>or= k+j|x>or=k) = P(x>or= k+j intersect x>or=k)/P(x>or=k) = P(x>or=k+j)/P(x>or=k) for j>or= 0 =...

Ok let me try this again... |(x2 + y2 - xy)/(x2 + y2)|<or= 1 because you are subtracting xy from the top so something smaller divided by something larger is a fraction less than one. Then, since the exponent is bounded by m=1, can you say f(x,y) is bounded by 0 and e, i.e. when you plug in...

|(x2 + y2 - xy)/(x2 + y2)| = |(r2 - rcosArsinA)/r2)| = |1-cosAsinA| which is bounded by 0 and 1 So f(x,y) is bounded by 1 and e. Is it enough to show that? Or do I need to use definition of limits... Also, for the second part, it cannot be extended because x=0=y is a singularity. Does...

Ohhhhhhhh! That is more simple than I thought it would be. I'm really sorry you had to hit me over the head with it for me to understand. But, I really do appreciate it. I know I am not very good at this.

Assume f(x) <or= g(x) <or= f(x)g(x). Since lim g(x) = 0, lim f(x) = lim f(x)g(x) = 0 by the Squeeze Theorem. Is that not correct logic then? I'm sorry if I seem to be processing this slow. I really am trying my best. I don't have a very strong background in Math.

Since f is bounded on the interval (c, infinity), f(x)g(x) will be bounded on the same interval. How do you know g(x) is bounded on it too? Assume f(x) <or= g(x) <or= f(x)g(x). Since lim g(x) = 0, lim f(x) = lim f(x)g(x) = 0 by the Squeeze Theorem.

Homework Statement The function exp[ (x2 + y2 - xy)/(x2 + y2) ] = f(x,y) is continuous on the open first quadrant. Prove it is bounded there. Prove f cannot be extended continuously to the closed first quadrant. The Attempt at a Solution Since f is a real-valued function...