Recent content by fateswarm

Well I can I show that for a negative 'a' F will not be approaching 1 for x->+oo. The integral of f(x) between 0 and +oo appears to be 'a' times the expected value of any Exp(a) so it's 1 for any 'a', hrm.. provided a>0 (for any Exp(a)) so I guess that might be enough. It's a bit confusing since...

I've got a bit of trouble. The derivative>0 appears to conclude to (a^2)xe^(-ax)>0 which appears to be true for any 'a' belonging in IR. It appears to be more direct to do 0<=F(x)<=1 which appears to conclude to 0<=(e^-ax)(1-ax)<=1 but how do I learn how to solve that inequality?

Oh, good idea (+thanks). I'll look into it. I may return if I see an issue.

Homework Statement Given the function F(x) = 1 - e^(-ax) - axe^(-ax) for x>=0 and 0 elsewhere, for which values of 'a' does the function constitute a CDF? Homework Equations The Attempt at a Solution I started with saying for that to occur, provided x1>x2 -> F(x1) > F(x2)...

The most intuitive way I found to look at it mathematically is to read the symbolism in reverse: e.g. P(y>max{X,Y}) for the serial problem. That way "y must be larger even than the larger of the two, hence both must be large enough [which is what I want in a serial situation]".

Thank you very much. I think I can go on from there; investigating it further.

What tells me that when I need to integrate it I can't just do (25-x^2)^0.5 and go on from there with common power of x integration? What is the thing that tells me "hang on there, this requires trigonometric integration"?

I think I figured it out, initially numerically and not intuitively, since I get the same result: It appears that for the PARALLEL problem one has to test the MIN between X, and Y rather than the max. That way the result more conveniently is, for a simpler example of "less than 1 hour"...

I currently wonder if P(a<max{X,Y}<b) is equivalent to P(a<X<b or a<Y<b) to begin with. They seem to answer two different questions. The first being "the probability of the one with the maximum value being between those two values" and the latter "the probability of either being between those...

How? Aren't X, Y restricted to be examined between the times 1 and 2? How could X, or Y go beyond those time frames?

On my studying material I noticed that for "devices in parallel" problems, a question of the type "Find the probability the device works for 1 year" is dealt with P(A or B) where P(A), P(B) are the probabilities of each of the instruments working for one year, which is something I intuitively...

It has no ambiguity, it is actually 1/(4-2). It was based on a Uniform distribution of U(2,4).

I had to do but it was given as a shortcut that and the result was How does one go from the 2nd to the 3d when there isn't an x^2 in the beginning there?

Ah, I got it, x is y^2.. not the other way around.

If a probability density function is fx(x) = 5x^4 for 0<x<1 and 0 elsewhere, then the expected value of E[x^0.5] is the integral of ∫x^0.5*fx(x) between 0 and 1 which is 10/11. However if I try to approach it as the expected value of Y=X^0.5 then I find fy(y) = 5*lny/y^4 and its expected...