Recent content by mathmajor23

Thanks a bunch! I, too, hate the annoying indicator functions, but my professor seems the need to "require" us to use them.

Ok, hopefully here's the final check: fx(x) = ∫ from x to infinity [2e^(-x-y)]dy =-2[e^(-x-y)] from y=x to y=∞ = 2e^(-2x) I[0,∞) (x) fy(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y) f(x|y) = 2e^(-x-y) / 2[e^(-y)-e^(-2y)] = [e^(-x)] / (1-e^(-y)) I[0,∞) (x) f(y|x) = 2e^(-x-y) /...

Let's start over. I originally had for the marginal of X: ∫ from 0 to x of [2e^(-x-y)]dy =-2[e^(-x-y)], evaluated from y=0 to y=x = 2[e^(-x)-e^(-2x)] I[0,∞) (x) To check this, I integrated this and it came out to 1, so I'm not sure why this is wrong.

Which part of f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x) is wrong?

Was my conditional of f(x|y) correct? Hmm is this the MD of x then? Integral from 0 to infinity [2e^(-x-y)dy] = -2(e^(-x-y)] from y=0 to y=infinity =2[1+e^(x)] I [0,infinity) (x)

I got that k=2, and checked it by seeing it integrated to 1. For the Marginal Distributions of X and Y, I got : f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x) f(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y) Then, for the conditionals, I got: f(x|y) = e^(-x) / [1-e^(-y)] I[0,infinity) (x)...

What's the integral of the inner part evaluate to? I tried using integration by substitution but it's not working.

Oops! So it should be ∫from 0 to ∞ ∫ from 0 to x [f(x,y)dxdy] ?

Yes, it's the same as single variable. I'm just unsure on my bounds of my integrals.

f(x,y) = ke^(-x-y), 0<x<∞ , 0<y<∞ , and x<y Find k to make this a PDF. So I set up: ∫(from 0 to ∞)∫(from 0 to ∞) [ke^(-x-y)]dxdy. Is this integral right? I also need to find the marginals of X and Y but my k has to be right first.

If you're getting ∞ as an answer, then the series diverges, not converges.

FX(1)(x) = P(X(1) <= x) = P(X1,...,Xn <= X) =1-P(X1,...,Xn > X) =1-P(X1>X)^n since the xi's are iid. =1-[1-P(X1 <= X)]^n =1-[1-F(x)]^n =1-[1-∫ from 0 to x (2tdt)]^n =1-(1-X^2)^n For the probability, P(X1>.2) = 1-Fx(1) (0.2) = (1-(0.2)^2)^n = (0.96)^n

P(A|A+B=c) = P(A|B=c-A) = P(A and B=c-A) / P(B=c-A) =P(α + β) / P(β) ?

I'm also trying to solve this similar problem, and also have no idea how to go about solving it.

Can anyone show me this problem step by step? I'm not picking up on any of this question, which is why I posted this.