Recent content by oyth94

Suppose the time in days until a component fails has the gamma distribution with alpha = 5, and theta = 1/10. When a component fails, it is immediately replaced by a new component. Use the central limit theorem to estimate the probability that 40 components will together be sufficient to last at...

How do I prove that given Xn converges almost surely to X, that f(Xn) will converge almost surely to f(X)?

Re: moment generating function question I'm not sure how you arrived at this answer..can you please explain?

Here is a proof question: For two random variables X and Y, we can define E(X|Y) to be the function of Y that satisfies E(Xg(X)) = E(E(X|Y)g(Y)) for any function g. Using this definition show that E(X1 + X2|Y) = E(X1|Y) + E(X2|Y) So what I did was I plugged into X = X1 + X2 E(E(X1 +...

Is there any other way to solve it besides using Catalan number and recursive relation and gamblers ruin?? Like using expected value or different distributions of some sort?

Hello .Sorry for all the questions I am asking...but I have another one I would like to seek help for.: Suppose a system has 10 components and that a particular time the jth component is working with probability 1/j for j= 1,2,...10. How many components do you expect to be working at that...

Thank you so much! I arrived with the same answer but I used either Poisson or Binomial. But my approach I didn't involve expected value...which I should...

Let X1,X2,…,Xn be independent random variables that all have the same distribution, let N be an independent non-negative integer valued random variable, and let SN:=X1+X2+⋯+XN. Find an expression for the moment generating function of SN so all i know is that it is i.i.d but i am not sure what...

There are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off, he turns it on. Initially all light bulbs are off. After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs...

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these...

okay so i forgot to mention i have to take the derivative after i factor/cancel out the t right. so then it becomes 1 + t/2! + t^2 /3! + ... derivative --> 1/2 + 2t/6 + ... and then after i take the derivative i set t=0 because mx1(0) = EX so from there i will get EX = 1/2 as you said before.

Thank you for your help. For how I did it was I used the series expansion for e^t which is the summation of (t)^k / k! so $$[(t^k / t!) - 1 ] / t$$ = [(1 + t/1 + t^2 / 2! + t^3 / 3! + ...) - 1] / t then the 1s cancel out and I can factor out the t so it becomes = 1 + t/2! + t^2 / 3! + ... so...

Suppose I have the MGF moment generating function mx(t) = (e^t -1)/t How can I find EX?

Hi I know this may be a silly question but i am doubting myself on how i did this question: Suppose X and Y are independent, with E(X) = 5 and E(Y) = 6. For each of the following variables Z, either compute E(Z) or explain why we cannot determine E(Z) from the available information: Z =...

Re: expected values: three fair coins Oh I see what you did! Thank you so much!