# MLE method for a statistical model with two random variables

Suppose for fixed $t>0$ we have two discrete random variables $X_t$ and $X_{t+1}$ where $X_t$ takes on $11$ values $$X_t\in\{0,1,2,\cdots, 10\}.$$ Also, we have $$P(X_{t+1}=X_t)=1-\lambda,\quad P(X_{t+1}=\min\{X_t+1,10\})=\lambda.$$ What is the ML estimator for $\lambda$?

The question is motivated by a model about bus engine replacement in microeconometrics. I have searched along this Wikipedia article and this lecture note. But I don't see how to do it since two random variables are involved. Could anyone come up with a theoretical reference for such an issue (MLE method for a statistical model with two random variables)? (Or this could be trivially reduced to the one-variable case?)

• Do you just have two random variables or several consisting of the complete histories of more than one bus engine with the values of $X_1, X_2,..., X_t$ ? That would also mean that we would assume that $t$ also takes on the discrete values $1, 2, 3, ..., t$ ? And does $X_1=0$ with probability 1 ?
– JimB
Nov 29 '15 at 5:01
• I'm now focusing on the case that $t$ is fixed. Thus there are only two random variables.
– Jack
Nov 29 '15 at 14:57

This is more of an extended comment than an answer.

I wonder if you meant to give conditional probabilities such that $P(X_{n+1}=x_n | X_n = x_n)=1-\lambda$ and $P(X_{n+1}=\min{(x_t+1,10)}|X_n=x_n)=\lambda$ with $P(X_1=0)=1$.

That way the likelihood of an observed history with $t=6$ of 0 0 1 1 2 3 would be $(1-\lambda)^2 \lambda^3$ with the maximum likelihood estimator being 3/5.

In general with an arbitrary value of $t$ and $m$ occurrences where there was a change from $X_i$ to $X_{i+1}$ the likelihood would be $(1-\lambda)^{t-m-1} \lambda^m$, the maximum likelihood estimator of $\lambda$ would be $m/(t-1)$ (except when $X_t=10$ as then $X_{t+1}$ must also equal 10 resulting in no information about $\lambda$).

But if you only know say two values $X_5$ and $X_6$, then you only have a sample size of 1 and the maximum likelihood estimator of $\lambda$ is 1 if $x_6 > x_5$ and 0 otherwise (unless $x_5=10$ when means that $x_6$ also equals 10 with probability 1 and for that case there is no information about $\lambda$).

Alternatively, if you have $n$ multiple sets of $X_t$ and $X_{t+1}$ (where $X_t < 10$ and if you can assume independence among the sets),then you simply have a binomial distribution and the maximum likelihood estimator is simply the number of observations of increases (m) divided by n.

I still think you need to give more specifics.