Autocorrelation of telegraph signal with Poisson Distribution,,,

Question

In the problem below, I'm not understanding how they came up with the formula for X(t) used to calculate the autocorrelation $E\Big[X(t)X(t+\tau)\Big]$. All they say about X(t) is that it has equal probability of switch polarity from +A to -A and vica versa... but, the picture they show of X(t), looks more like its a discrete signal X[n] instead of a continous signal X(t)....

Answer 1

On base of the fact that $X\left(t\right)$ and $X\left(t+\tau\right)$ only take values in $\left\{ A,-A\right\} $ we are allowed to conclude that: $$\Omega=\left\{ X\left(t\right)=X\left(t+\tau\right)\right\} \cup\left\{ X\left(t\right)\neq X\left(t+\tau\right)\right\} $$

Moreover: $$\left\{ X\left(t\right)=X\left(t+\tau\right)\right\} \cap\left\{ X\left(t\right)\neq X\left(t+\tau\right)\right\} =\varnothing$$

This justifies the equality: $$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\right]=$$$$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\mid X\left(t\right)=X\left(t+\tau\right)\right]P\left(X\left(t\right)=X\left(t+\tau\right)\right)+$$$$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\mid X\left(t\right)\neq X\left(t+\tau\right)\right]P\left(X\left(t\right)\neq X\left(t+\tau\right)\right)$$

Under condition $X\left(t\right)=X\left(t+\tau\right)$ we find that $X\left(t\right)X\left(t+\tau\right)=A^{2}$.

Under condition $X\left(t\right)\neq X\left(t+\tau\right)$ we have $X\left(t\right)X\left(t+\tau\right)=-A^{2}$.

So we can extend the equality with: $$\cdots=A^{2}P\left(X\left(t\right)=X\left(t+\tau\right)\right)+\left(-A^{2}\right)P\left(X\left(t\right)\neq X\left(t+\tau\right)\right)$$

Answer 2

\begin{align} E[X(t) X(t+ \tau)] &= E[X(t) X(t+ \tau)| X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal} ] P(X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal}) + \\ &+ E[X(t) X(t+ \tau)| X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals} ] P(X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals}) \\ &= A^2 P(X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal}) - \\ &-A^2 P(X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals}) \\ &= A^2 P(\textrm{number of crossings is even}) - \\ &-A^2 P(\textrm{number of crossings is odd}) \\ \end{align}