# Geometric distribution from exponential estimation

Problem:

$$X \sim \exp(\lambda)$$

Now $$X$$ is descretized to $$Y$$ using floor function

$$Y_1, Y_2, \dots,Y_n$$ is available from $$Y$$ distribution

I need to find method of moment estimator for $$\lambda$$ using $$Y$$ sample value

My solution:

$$Y \sim$$ geometric distribution with parameter $$p$$

Here $$p=1-\exp(-\lambda)$$

Now, moment estimator for $$p$$ is

$$\hat{p} = \frac{1}{\overline{Y}}.$$

If we keep this $$p$$ value in above equation to get $$\lambda$$. $$\lambda= -\ln\left(1-(1/\overline{Y})\right)$$

$$\lambda= \ln\left(1+(1/\overline{Y})\right)$$
The distribution of $$\lfloor X \rfloor$$ when $$X$$ has pdf $$\lambda e^{-\lambda x}$$ (which I assume is the convention you are using for the exponential distribution) will have PMF $$P(Y=y)=\int_y^{y+1} \lambda e^{-\lambda x} \, dx = -e^{-\lambda(y+1)}+e^{-\lambda y}=e^{-\lambda y} \left ( 1-e^{-\lambda} \right )$$. Thus it is geometric with success probability parameter $$p=e^{-\lambda}$$, under the failure-counting convention (so $$0$$ is a possible value).
The mean of it is $$\frac{1}{p}-1=e^\lambda-1$$, so $$\overline{Y}$$ is a consistent and unbiased estimator for $$e^\lambda-1$$. Therefore $$\log(1+\overline{Y})$$ is a consistent but probably biased estimator for $$\lambda$$.
If you use $$\hat{p}$$ you will arrive at an algebraically equivalent expression, by using $$\frac{1}{\hat{p}}-1=\overline{Y}$$ and $$\hat{p}=e^{-\hat{\lambda}}$$ and solving for $$\hat{\lambda}$$.