# Particle relocating at constant rate with Poisson troubles

Consider a fragment beginning at the the beginning and also relocating along the favorable actual line at a constant rate of 1. Intend there is a counter which clicks randomly time periods adhering to the exponential distribution with parameter $\lambda$ and also whenever the counter clicks, the placement $x > 0$ of the fragment back then immediately transforms to the placement $x/2$. We desire to compute the anticipated ordinary rate of the fragment.

I do not actually have any kind of suggestion of just how to deal with addressing this. Below are a number of relevant troubles which appear a lot more hard to me :

- Modify the puzzle to make sure that when the counter clicks, the fragment relocates from $x$ to a factor picked evenly randomly from $[0,x]$.
- The fragment begins relocating as over yet whenever the counter clicks, its rate rises by 1 (the first rate was 1). What is the predicted time when the fragment strikes the placement 1? What is the anticipated rate when the fragment strikes the placement 1?

This is not a research trouble. Any kind of remedies, tips, ideas will certainly be valued.

Many thanks,

Just for the document, I such as Shai Covo is address far better. Yet the OP asked me to upload my remedy too, so below it is.

Allow $X_t$ be the placement of the object sometimes $t$. Offered $N$ clicks in $[0,t]$, allow $\tau_1, \tau_2, \ldots \tau_N$ be the moments of those clicks. Allow $T_i$ be the $i$th interarrival time, to make sure that $T_1 = \tau_1$, $T_{N+1} = t - \tau_N$, and also $T_i = \tau_i - \tau_{i-1}$ or else. Hence $t = \sum_{i=1}^{N+1} T_i$.

By buildings of the rapid circulation, $E[T_i|N] = E[T_j|N]$ for all $i, j$. Hence $$t = \sum_{i=1}^{N+1} E[T_i|N] = E[T_i|N] (N+1) \Rightarrow E[T_i|N] = \frac{t}{N+1}.$$

If $N=0$, after that $X_t = T_1$. If $N = 1$, $X_t = \frac{1}{2}T_1 + T_2$. If $N = 2$, $X_t = \frac{1}{4}T_1 + \frac{1}{2}T_2 + T_3$, and also, as a whole, $X_t = \sum_{i=0}^N \frac{T_{N+1-i}}{2^i} $. Hence $$E[X_t|N] = \sum_{i=0}^N \frac{E[T_{N+1-i}|N]}{2^i} = \frac{t}{N+1}\left(2 - \frac{1}{2^N}\right).$$

Since $E[X_t] = E[E[X_t|N]]$, we simply need to compute $E\left[\frac{1}{N+1}\right]$ and also $E\left[\frac{1}{(N+1)2^N}\right]$. Given that $N$ is Poisson$(\lambda t)$, we have $$E\left[\frac{1}{N+1}\right] = \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} e^{-\lambda t}}{(n+1) n!} = e^{-\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} }{(n+1)!} = \frac{e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n+1} }{(n+1)!} $$ $$= \frac{e^{-\lambda t}}{\lambda t} \left(\sum_{n=0}^{\infty} \frac{(\lambda t)^n }{n!} - 1 \right) = \frac{e^{-\lambda t}}{\lambda t} \left(e^{\lambda t} - 1 \right) = \frac{1}{\lambda t} - \frac{e^{-\lambda t}}{\lambda t}.$$

Similarly, $$E\left[\frac{1}{(N+1)2^N}\right] = \frac{2e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\frac{\lambda t}{2})^{n+1} }{(n+1)!} = \frac{2e^{-\lambda t}}{\lambda t} \left(e^{\lambda t/2} - 1 \right) = \frac{2e^{-\lambda t/2}}{\lambda t} - \frac{2e^{-\lambda t}}{\lambda t}.$$

Therefore,

$$E[X_t] = E\left[\frac{2t}{N+1}\right] - E\left[\frac{t}{(N+1)2^N}\right] = \frac{2}{\lambda}\left(1 - e^{-\lambda t/2}\right),$$ which is specifically what you get if you address the differential formula in Shai Covo is solution.

So the anticipated ordinary rate (rate, in fact, given that the ordinary rate is practically 1) is $$\frac{2}{\lambda t}\left(1 - e^{-\lambda t/2}\right).$$

Sketch of remedy : Let $X_t$ represent the area of the fragment sometimes $t$, and also set $f(t) = {\rm E}(X_t)$. Take into consideration the moment $t + \Delta t$, $\Delta t \approx 0+$. With probability of concerning $1 - \lambda \Delta t$ the fragment remains to place $X_t + \Delta t$, whereas with probability of concerning $\lambda \Delta t$ it relocates to placement of concerning $X_t/2$. This leads straight to a primary differential formula in regards to $f(t)$, where you get $f(t)$. The anticipated ordinary rate is after that $f(t)/t$.