I know this is probably very simple but I can't seem to solve this correctly.
X transitions to Y with a probability of 0.75, X transitions to X with a probability of 0.25 and X does not transition to Z.
Y transitions to X with a probability of 0.2, Y transitions to Y with a probability of 0.6, and Y transitions to Z with a probability of 0.2
Z transitions to Y with a probability of 0.2 and Z transitions to Z with a probability of 0.8
These probabilities render a stationary distribution with the following system of linear equations:
$$P_\infty(X) = \frac{1}{4}P_\infty(X) + \frac{1}{5}P_\infty(Y)$$ $$P_\infty(Z) = \frac{4}{5}P_\infty(Z) + \frac{1}{5}P_\infty(Y)$$ $$P_\infty(Y) = \frac{1}{5}P_\infty(Z) + \frac{3}{4}P_\infty(X) + \frac{3}{5}P_\infty(Y)$$
I can reduce the second equation to $\frac{1}{5}P_\infty(Z) = \frac{1}{5}P_\infty(Y) \rightarrow P_\infty(Z) = P_\infty(Y)$
I can reduce the first equation to $\frac{3}{4}P_\infty(X) = \frac{1}{5}P_\infty(Y) \rightarrow P_\infty(X) = \frac{4}{15}P_\infty(Y)$
I can rewrite the third equation as $\frac{1}{5}P_\infty(Y) + \frac{1}{5}P_\infty(Y) + \frac{3}{5}P_\infty(Y) = P_\infty(Y)$
However this leads me to $P_\infty(Y) = \frac{5}{5}P_\infty(Y)$, or $P_\infty(Y) = P_\infty(Y)$
Where did I go wrong?
There are no computational errors. The important observation is that writing those equations always results in a system that is rank-deficient. One equation is therefore replaced by the condition that the sum of all stationary probabilities should equal $1$. If you do that, you'll get $$ \left[\frac{2}{17}, \frac{15}{34}, \frac{15}{34} \right] $$ as solution, where indeed $P_\infty(Y) = P_\infty(Z)$.
Why are those equations always linearly dependent? If you write them in matrix form, you get $(Q-I)\,P_\infty = 0$, where $Q$ is the transition matrix of your Markov chain. Since $Q$ is a stochastic matrix, all its columns add to $1$. Subtracting the identity makes that sum $0$, which means that there exists a linear combination of the rows of $Q-I$ (namely, with unit coefficients) that is the $0$ vector. Hence $Q-I$ is not full rank.
Intuitively, the equations in $(Q-I)\,P_\infty = 0$ express flow conservation, and they say nothing about the amount of "flow." You need an additional equation to say that the "flow" is a probability distribution.