Recent content by shan

Not to hijack the topic but your explanation was so good and I was also having problems with formal definitions; what is the difference between limits and convergence? It seems to me like you can use them interchangeably like this: The sequence Sn converges to L as n approaches infinity. The...

If anyone was interested, the answer was no. The counter example was: Let P(X_1 = 1) = P(X_1 = -1) = 0.5. Let X_2 = -X_1 and Y_1 = Y_2 = X_1. X_1 X_2 = -X_1^2 = -1 almost surely, but Y_1 Y_2 = X_1^2 = 1

If anyone was interested: Say h(Y_n) = Z_n, h(Y) = Z E(g(Z_n)) \rightarrow E(g(Z)) for every g that is bounded and continuous (from definition) E(f(Y_n)) \rightarrow E(f(Y)) for every f that is bounded and continuous (from definition) E(g(h(Y_n)) \rightarrow E(g(h(Y)) is true because...

Ah sorry, it's the stuff I put in the brackets ie. X1 ~ Y1 : the distribution of X1 has the same distribution of Y1, X1 and Y1 being random variables. The same with X2 ~ Y2 and X1X2 ~ Y1Y2.

Homework Statement If X1 ~ Y1 and X2 ~ Y2, then X1X2 ~ Y1Y2, prove or find a counterexample. (the distribution of X1 has the same distribution of Y1 etc) Homework Equations - The Attempt at a Solution I'm guessing the statement is true. For example if X1 and Y1 were both uniform...

Given the definition: For real-valued random variables X_n, n\geq1 and X, then X_n\stackrel{D}{\rightarrow}X if for every bounded continuous function g: R \rightarrow R, E_n[g(X_n)]\rightarrow E[g(X)] I want to prove the continuous mapping theorem: If X_n\stackrel{D}{\rightarrow}X then...

I just want to check my answers/reasoning as I'm not sure if I assumed the right things to do these problems. N = {N(t), t>=0} ~ Poisson(1) and N_{(t,t+h]} = N(t+h)-N(t) Determine P(N(4) =3|N(2) = 1) Here I presumed that since N(2) = 1, then there must be 2 more arrivals in the interval...

Thank you very much for your help :)

whoops sorry, typo :blushing: that first part is \int_0^x \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{12} which shows I don't really understand what I'm doing but now that you mentioned it... is the second part then given by \int_0^2 \frac{1}{6}u du + \int_2^x \frac{1}{3}...

I've gotten a weird answer after doing the problem but I'm stuck as to where I messed up. The density function is this: f_{X} (x) = \frac{1}{6}x for 0<x\leq2 = \frac{1}{3}(2x-3) for 2<x<3 and 0 otherwise And the question is to find the distribution function. So integrating for the...

ok, thanks for your help alphanumeric, it's a little easier to see why it is true that way.

oops sorry, I missed out a 20 in the integral so I'm asking if \int 20 d\theta = 20\theta I have never heard of this formula? And also, AlphaNumeric, could you explain why you suggested i = e^{\frac{i\pi}{2}+2\pi ni}??

The first one, integration, I just want to check my answer. \int \frac{1}{64} (\cos6\theta + 6\cos4\theta + 15\cos2\theta + 20) = \frac{1}{64} (\frac{\sin6\theta}{6} + \frac{6\sin4\theta}{4} + \frac{15\sin2\theta}{2} + 20\theta + c I just wasn't sure if the integral of a constant wrt theta...

I'm not too sure where to post this so feel free to move it :) Anyway I'm hoping someone could explain the answer of this problem to me (I would ask my lecturer but he's conveniently away for the week for a meeting). Suppose X and Y are iid continuous random variables with density f...

oops sorry, I'm missing the lim as m->infinity signs... well if you put those in, it makes sense :)