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 Nov 26 comment Simple Renewal Process Question Only a bit. As I gathered, the distribution of the total lifetime of the component $C_t$ can be thought of as the sum of the distributions of the current age $A_t$ and the residual life $B_t$, but I don't know much of anything else regarding this matter. I should add that I'm going to bed now (I've got class in 4 hours) so if you have a formal answer to my question, I'd appreciate it if you posted it and I can ask questions about it tomorrow if I have any. Thanks for your time! Nov 26 comment Simple Renewal Process Question Yes, I have seen that in class but only briefly. How can I apply that to this problem? Nov 26 comment Simple Renewal Process Question Ah, you do make a good point. In that case, I'm thinking that the distribution of $A+B$ would be uniform with $E(A+B)=150$. Am I right in thinking that? Nov 12 comment Optional Sampling Theorem Application on a Martingale Thanks for your response. I actually made a typo in my original post. $M_n = [(1-p)/p]^{S_n}$, not $[(1-p)p]^{S_n}$. This, however, shouldn't change your answer (aside from changing your $p(1-p)$ terms to $(1-p)/p$). Either way, this makes perfect sense. Thank you! Nov 5 comment Conditional Expectation Die Roll Thanks. Is there any way to arrive at $y/2$ given the work I have in my original post? Nov 5 comment Conditional Expectation Die Roll Doh, why didn't I think of that? Thanks! Nov 4 comment Conditional Expectation Die Roll Sorry for not being clear enough. What @MichaelHardy said is correct. I need to find the expected value of the first roll given the sum of the two dice. Oct 10 comment Poisson Process Arrival Probability It's no question why you have so much reputation. Thank you. Oct 10 comment Poisson Process Arrival Probability Well, the homework's not graded and is solely for our benefit of understanding the material, so it wouldn't hurt. Also, it's 2 am here and I need sleep so I'll go with yes, please give an answer. Oct 10 comment Poisson Process Arrival Probability Sorry, but I'm not really seeing where this is going. Care to elaborate a bit? Oct 10 comment Poisson Process Arrival Probability I know that the expected value of the waiting time is $1/{\lambda}$. Is that what you're getting at? Oct 10 comment Branching Process Extinction Probability This is great, thanks! Oct 10 comment Branching Process Extinction Probability Scratch that--A little consideration of my own helped me understand it. Thanks! Oct 10 comment Branching Process Extinction Probability This all makes sense up until you write up your equation. I can see how to apply it, but I'm not seeing why $$P(X_2 = 0, X_1 > 0) = P(X_2 = 0) - P(X_1 = 0).$$ Could you explain that? Sep 24 comment Markov Chain Reach One State Before Another Thanks, Byron. That's exactly what I was looking for. Incidentally I was able to figure it out when I needed it, but thanks anyway! Aug 19 comment Create a C++ program to evaluate the following series: $\sin x \approx x - \frac{x^3}{3! }+\frac{x^5}{5!}-\frac{x^7}{7!}\cdots\pm\frac{x^n}{n!}$ You'd change it to \approx because the series is approximating sin(x), not evaluating it exactly. Aug 18 comment Random Walk Expected Number of Visits Ah okay, that makes sense. Thanks for your help! Aug 18 comment Random Walk Expected Number of Visits So if it were the case that both A and E were destination states, the answer to the question: Suppose a walker starts in vertex C. What is the expected number of visits to B before the walker reaches A? would come from the same matrix as the question: Suppose a walker starts in vertex C. What is the expected number of visits to B before the walker reaches E? Aug 18 comment Random Walk Expected Number of Visits First, you are absolutely right about the minus signs. I somehow convinced myself that $$0 - \frac{1}{3} = \frac{1}{3}.$$ And to be clear, that (C, B)th entry of the matrix M accounts for the fact that our destination is vertex A, right? And that would be because we designated A as an absorbing state? Aug 13 comment Invariant Probability Vector Yep, that's exactly what I know how to do. You've been very helpful; thanks again!