I have the following problem, in a statement I am given the following conditional probabilities

$$P(x_i | x_{i-1}) = 0.7$$

$$P( \overline{x_i} | x_{i-1}) = 0.3$$

$$P(x_i | \overline{x_{i-1}}) = 0.4$$

$$P(\overline{x_i} | \overline{x_{i-1}}) = 0.6$$

These indicate the probability of an event occurring or not occurring given whether or not it occurred the day before. Based on that, the following tree of probabilities is made up until a day $i=3$, but in theory it would be up to a day $i=n$

tree of probabilities here

We are asked to express the probability of the event occurring on a day $i$ given that it did not occur today, i.e.

$$P(x_i | \overline{x_0})$$

I can't find an expression for what is required for any day $i$ in the future. Until day two, based on the probability tree, it is clear that there would be two possible paths and the probability would be

$$P( x_2 | \overline{x_{1}}) P(\overline{x_{1}}|\overline{x_0}) + P(x_2 | x_{1}) P(x_{1}|\overline{x_0})$$

$$ 0.6 \cdot 0.4 + 0.4 \cdot 0.7 $$

But when generalizing and finding an expression for a day $i>2$, since we haven't seen the random variable theory, I don't know how to do it.


Describe the problem using the tools of Linear Algebra and Matrices.

Let $A=\begin{bmatrix}0.7&0.4\\0.3&0.6\end{bmatrix}$ and let $v_n=\begin{bmatrix}P(x_n)\\P(\overline{x_n})\end{bmatrix}$

We have $v_{n}=Av_{n-1}=A^nv_0$ and in particular $P(x_n\mid \overline{x_0})$ will be the first entry of $A^n\begin{bmatrix}0\\1\end{bmatrix}$.

Armed with this set up, one can now apply all the usual tools and methods available from Linear algebra, in particular the use of eigenvalues and of diagonalization.

By diagonalizing $A$ as $SDS^{-1}$ with $D$ diagonal, it follows that $A^n = SD^nS^{-1}$ at which point we can simplify $A^n v_0$. I leave the details to you to work out.

  • $\begingroup$ Thank you very much. I would never have thought of anything like that. $\endgroup$
    – Fmkit
    May 27 '20 at 0:45

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