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I came across a type of random process earlier with the structure $$ X(t) = (-1)^{N(t)}Y,\ t\geqslant 0 $$ where $\{N(t):t\geqslant 0\}$ is a homogeneous Poisson process with intensity $\lambda$ and $Y$ is a symmetric Rademacher random variable, i.e. $\mathbb P(Y=1)=1/2 = \mathbb P(Y=-1)$ independent of $\{N(t)\}$. It was called a random telegraph signal. Obviously $X(t)=\pm 1$ as according to the number of jumps in the Poisson process and $Y$ determines the value of $X(0)$.

Searching for "random telegraph signal," I could not find anything related to this type of process. I'm interested in literature (textbooks or papers) that study this process and its generalizations in more detail, but I do not know what to search for.

Math1000
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1 Answers1

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I'm trying to work out the pmf of X(t) here:

P[X(t) = y |Y = y]

= P[N(t) is even]

= $\sum_{j=0}^{\infty} \frac{{(\lambda t)}^{2j}}{(2j)!} e^{-\lambda t} $

= $e^{-\lambda t} \frac{1}{2}\{e^{\lambda t}+e^{-\lambda t}\}$

=$\frac{1}{2}\{1+e^{-2\lambda t}\}$

Similarly,

P[X(t) = -y |Y = y]

= P[N(t) is odd]

= $\sum_{j=0}^{\infty} \frac{{(\lambda t)}^{2j+1}}{(2j+1)!} e^{-\lambda t} $

= $e^{-\lambda t} \frac{1}{2}\{e^{\lambda t}-e^{-\lambda t}\}$

=$\frac{1}{2}\{1-e^{-2\lambda t}\}$

Therefore,

P[X(t) = 1]

=P[X(t) = 1|Y=1]P[Y=1] + P[X(t) = 1|Y=-1]P[Y=-1]

=$\frac{1}{2}\{1+e^{-2\lambda t}\} \frac{1}{2}$ +$\frac{1}{2}\{1-e^{-2\lambda t}\}\frac{1}{2}$

=$\frac{1}{2}$

P[X(t) =-1]

=P[X(t) = -1|Y=-1]P[Y=-1] + P[X(t) = -1|Y=1]P[Y=1]

=$\frac{1}{2}\{1+e^{-2\lambda t}\} \frac{1}{2}$ +$\frac{1}{2}\{1-e^{-2\lambda t}\}\frac{1}{2}$

=$\frac{1}{2}$

Thus X(t) is equally likely to be 1 or -1 at anytime.

This is essentially the same to the signal discussed in $\pmb{Example\ 6.22\ Page\ 350}$ of the textbook ${Probability\ and\ random\ process\ for\ electrical\ engineering,\ second\ version}$ by $A.\ Leon-Garcia$.

Hope this helps.

  • This doesn't really answer the question, but indeed this is the result for the limiting behaviour of $X(t)$. Try computing $X(t)$ conditioned on ${Y=1}$ for some finite $t$, by comparison. – Math1000 Jun 17 '24 at 09:22
  • I tried googling "random telegraph signal" and it yields a lot of results talking about random telegraph signal noise in semiconductor devices. – B. Highwood Jun 17 '24 at 20:47