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visits member for 2 years, 5 months
seen Oct 20 at 15:06

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!