MLE for $p$ in Geometric distribution from Exponential distribution (two methods, two results)

Question

Let $Y_n$ given as $\mathrm{ceil}(X_n)$, where $\mathrm{ceil}(x):=$ the least integer greater than or equal to $x$ and $(X_n)$ is a sequence of iid random variables from $\mathrm{Exp}(\theta),~\theta>0$.

Then $Y_n\sim Geo(p)$, where $p=p(\theta):=1-e^{-\theta}$ and since the maximum likelihood estimator (mle) for $\theta$ is given as $\frac{1}{\overline{X}}$, the mle for $p(\theta)$ is $1-e^{-\frac{1}{\overline{X}}}$.

If we compute directly the mle for $p$ using $\mathbb{P}(Y_1=y)=(1-p)^{y-1}p$ for $y\in \mathbb{N}$, we get that the maximum likelihood estimator for $p$ is given as $\frac{1}{\overline{Y}}=\frac{1}{\overline{\mathrm{ceil}(X)}} $, which is not the same as the previous result.

Is there some contradiction in these two results, or some fallacy?

Thank for the help.

Answer 1

As Math-fun says, you are in effect using two different sets of information, one unrounded and the other rounded up, so you should not expect the same result.

For example,

• if you see the data $0.1, 0.2, 0.3, 1.1, 2.1$ then $1-e^{-1/\overline{x}} \approx 0.732$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$.
• if you see the data $0.7, 0.8, 0.9, 1.9, 2.9$ then $1-e^{-1/\overline{x}} \approx 0.501$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$ again since you round up to the same integers.

Here is some R code to simulate a thousand samples of size $10$. From the chart below you can reasonably conclude:

• the two maximum likelihood estimators for $p$ give slightly different estimates in practice, largely due to the particular sample data effects on the two expressions
• the estimators are usually closer to each other than they are to the actual parameter used in the simulation ($0.3$ in this case)
• the estimator using the ceiling function will only give discrete values; in retrospect this is obvious looking at the formula, as it can only be the sample size divided by an integer

.

set.seed(2021)
n <- 10
p <- 0.3
theta <- -log(1-p)
Xdat <- matrix(rexp(n*10^3, rate=theta), ncol=n)
MLE_p_exp <- 1 - exp(-1/rowMeans(Xdat))
MLE_p_geo <- 1 / rowMeans(ceiling(Xdat))
plot(MLE_p_exp, MLE_p_geo, xlim=c(0,0.8), ylim=c(0,0.8), pch=3)
abline(0,1, col="red")