I am confusing myself with a problem which I give below. Can someone please comment and point out my mistakes.
Say I have a two state system $S_1$ and $S_2$ the probability to transition to the two state (starting from state $S_1$) is $P_1$ and $P_2$. Note $P_1 + P_2 =1$.
In order to model this I assume that there are two independent random numbers $r_1$ and $r_2$ following standard normal distribution $N(0,1)$ that trigger the transition i.e. I create a random number $r$ defined as:
$r = a\times r_1 + \sqrt{(1-a^2)} \times r_2$
where both $r_1, r_2 \sim N(0,1)$ and $a$ is some correlation. For the two state system whenever $P(r> r_0) > P_2$ we have a transition to the state $S_2$ or else we stay at state $S_1$.
We can also compute the probability of transition to state $S_2$ conditional on generating a random number $r_1$. This conditional probability can be computed as follows:
$P(transition \ to \ S_2| r_1) = N(\frac{N^{-1}(P_2) -a\times r_1}{\sqrt{(1-a^2)}} ) $
and
$P(staying \ at \ S_1| r_1) = 1 - P(transition \ to \ S_2| r_1)$.
The formulation takes care of the normalization of the conditional probabilities i.e
$P(staying \ at \ S_1| r_1) + P(transition \ to \ S_2| r_1) = 1$
Now say I have a 3 state system (and the same two underlying random variable triggering the transition) for which I can compute:
$P(transition \ to \ S_2| r_1) = N(\frac{N^{-1}(P_2) -a\times r_1}{\sqrt{(1-a^2)}} ) $
$P(transition \ to \ S_3| r_1) = N(\frac{N^{-1}(P_3) -a\times r_1}{\sqrt{(1-a^2)}} ) $
and
$P(staying \ at \ S_1| r_1) = 1 - P(transition \ to \ S_2| r_1) - P(transition \ to \ S_3| r_1)$.
With this 3 state situation I am running into computational and intuitive issue which I explain below now:
Say I have a small $P_2 = P_3 = 0.45$ i.e. high probability of transition and probability to stay at $S_1 = 0.1$.
Also say I am trying to to compute $P(transition \ to \ S_2| r_1)$ and $P(transition \ to \ S_3| r_1)$ for a very small $r_1=-5$. Let us also choose $a=0.5$.
Then based on the way I am computing $P(transition \ to \ S_2| r_1)$ and $P(transition \ to \ S_3| r_1)$ above I get
$P(transition \ to \ S_2| (-5)) = N(\frac{N^{-1}(0.45) -a\times (-5)}{\sqrt{(1-a^2)}} ) = N(3.165) = 0.9992$
$P(transition \ to \ S_3| (-5)) = N(\frac{N^{-1}(0.45) -a\times (-5)}{\sqrt{(1-a^2)}} ) = N(3.165) = 0.9992$
Now you see that I made a reasoning mistake somewhere which is causing
$P(transition \ to \ S_2| (-5)) + P(transition \ to \ S_3| (-5)) > 1$
and I am unable to figure out my mistake.
It would be very kind if someone can comment. Thanks in advance.
Below is a solution to your problem as I understand it. Your main mistake was that you only considered one threshold when dealing with more than 2 states. For the sake of readability I start with the 2 states problem and then address the general case with $n$ states.
Let $S_1,S_2$ be two states, $P_1$ the probability of staying in $S_1$ and $P_2$ the probability of moving from $S_1$ to $S_2$ (with $P_1 + P_2 = 1$). In the following we assume that we are in the state $S_1$.
We define $$r=a\times r_1 + \sqrt{(1-a^2)}\times r_2$$ where $r_1$ and $r_2$ are two random variables which follow a standard normal distribution $N(0,1)$. Note that, since $a$ is a constant, $r$ also follows a standard normal distribution $N(0,1)$.
Let $r_0$ be the threshold such that $$P(r \leq r_0) = P_1 \text{ and } P(r > r_0) = P_2\text{.}$$ Note that $r_0$ is the solution of the equation $\Phi(x) = P_1$ where $\Phi(x)$ is the cumulative distribution function of the standard normal distribution.
Let $b$ be a constant. We can now compute \begin{align} P\left(\left.\text{transition to $S_2$} \right| r_1=b\right) &= P\left(\left.r > r_0 \right| r_1=b\right) \\ &= P\left(\left.a\times r_1 + \sqrt{(1-a^2)}\times r_2 > r_0 \right| r_1=b\right) \\ &= P\left(\left.r_2> \frac{r_0 - a\times r_1}{\sqrt{(1-a^2)}} \right| r_1=b\right) \\ &= P\left(r_2> \frac{r_0 - a\times b}{\sqrt{(1-a^2)}}\right) \\ &= 1- P\left(r_2 \leq \frac{r_0 - a\times b}{\sqrt{(1-a^2)}}\right) \\ &= 1- \Phi\left(\frac{r_0 - a\times b}{\sqrt{(1-a^2)}}\right)\text{.} \end{align}
We now assume that we have $n$ states $S_1, S_2, \ldots, S_n$. Let $P_i$ be the probability of moving from $S_1$ to $S_i$, such that $\sum_{i = 1}^n P_i = 1$.
We define $r$ as above and we are now looking for several thresholds $r^{i}_0$ with $1\leq i\leq n-1$ such that $$P(r \leq r^{1}_0) = P_1 \text{, } P(r^{i-1}_0 < r \leq r^{i}_0) = P_i \text{ and } P(r^{n-1}_0 < r) = P_n.$$ Note that $r^{1}_0$ is the solution of the equation $\Phi(x) = P_1$ where $\Phi(x)$ is the cumulative distribution function of the standard normal distribution and the other thresholds can be similarly computed by first noticing that $P(r^{i-1}_0 < r \leq r^{i}_0) = P(r \leq r^{i}_0) - P(r \leq r^{i-1}_0)$.
Let $b$ be a constant. We can now compute \begin{align} P\left(\left.\text{transition to $S_i$} \right| r_1=b\right) &= P\left(\left. r^{i-1}_0 < r \leq r^{i}_0 \right| r_1=b\right) \\ &= P\left(\left. r^{i-1}_0 < a\times r_1 + \sqrt{(1-a^2)}\times r_2 \leq r^{i}_0 \right| r_1=b\right) \\ &= P\left(\left. \frac{r^{i-1}_0 - a\times r_1}{\sqrt{(1-a^2)}} < r_2 \leq \frac{r^{i}_0 - a\times r_1}{\sqrt{(1-a^2)}} \right| r_1=b\right) \\ &= P\left( \frac{r^{i-1}_0 - a\times b}{\sqrt{(1-a^2)}} < r_2 \leq \frac{r^{i}_0 - a\times b}{\sqrt{(1-a^2)}} \right) \\ &= P\left( r_2 \leq \frac{r^{i}_0 - a\times b}{\sqrt{(1-a^2)}} \right) - P\left( r_2 \leq \frac{r^{i-1}_0 - a\times b}{\sqrt{(1-a^2)}} \right) \\ &= \Phi\left(\frac{r^{i}_0 - a\times b}{\sqrt{(1-a^2)}}\right) - \Phi\left(\frac{r^{i-1}_0 - a\times b}{\sqrt{(1-a^2)}}\right)\text{.} \end{align}