Recent content by kwy

Many thanks.

Got it. Thanks heaps.

Hi Guys, I think the means may be of use somewhere. There is no 2nd part to the question, the aim is to determine values of a that will i) minimise VarZ and ii) maximise VarZ. Following differentiation, a = sigma^2/(1+sigma^2). From this value and 2 random variables, how can I check the...

Thanks so much Ray and Stephen. It makes sense now.

Hi Ray, since the RVs X and Y are independent, I don't have to worry about the Cov(XY) right? Hence, the equation is now VarZ = a^2VarX + (1-a)^2VarY (VarX = 1, VarY = sigma^2) = a^2 + (1-a)^2*sigma^2 and where do the mean values come in? How can I differentiate this? Thanks in...

Thanks for your patience, the course went so fast, I am still trying to revise all the notes between full time work. so Var(W)= k^2Var(P) does Var(W+Q)= VarW + VarQ + 2xCov(W,Q)??

Yes, they are independent random variables. Can you please provide a bit more guidence? I just know Var(W) = kEX^2 -(kEX)^2? Am I totally off track?

Homework Statement EX=EY=5, VarX=1, VarY=sigma^2 >1 Z=aX+(1-a)Y, 0<=a<=1 find a that minimizes VarZ, and another a that maximize VarZ Homework Equations The Attempt at a Solution Not even sure where to begin *EX=5, VarX=1 thus EX^2 = 26 marginal px(x) =26/x^2 = 5/x but...

Hi aim1732 Thank you for your help. It was very rude of me to leave it so late. The assignment has already been marked, and that question was not even included in the marking (only half of the questions are randomly chosen for marking). Thanks again. Cheers kwy

Sorry, I'm going to try using LaTex to see if it makes more sense. If the lines below does not make sense, I apologise. Would you mind letting me know where you had to use L'Hopital Rule, I cannot work out where this can be applied: My previous workings were: 2(0 + \int 1/u * 1/4u-2 du)...

Thanks aim1732, but I'm still a little off. diff lnu = 1/u, int (2u-1)^-2 = -1/2(2u-1) Thus equation is now: 1) 2([-lnu/2(2u-1)] u=infinity to 1 + int(1/u)du + int(1/4u-2)du) 2) 2( 0 + int(1/u)du + int(1/4u-2)du) 3) 2([-1/u^2] + [ln(4u-2)]) where...

Homework Statement integrate (x*2e^x)/(2e^x-1)2 from x=0 to infinity Homework Equations The Attempt at a Solution let t=2e^x-1 => x=ln((t+1)/2) dt = 2e^x dx Thus equation is now integrate (ln((t+1)/2))/t^2 dt from t=1 to infinity Then let u = (t+1)/2 => 2du=dt Equation now...

I got it. You are right about line 6 where the summation's lower bound should y=1. This means the summation part is: sum(y=1 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 which equates to say A - B where A = sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 = 1 B where y is 0 giving...

Hi I like Serena The question on the sheet definitely says 1 - (1-p)^n+1 ----------------- p(n+1) Cheers and thanks.

Oh dear, if EY is p(n+1), I think I somehow got the entire equation to equal to 1. I believe I went off the wrong track after line 3. Would you agree? Sorry, it's been close to 20 years since I've done this kind of maths. My brain just cannot interpret things as fast. I'm really...