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 $\lambda$. $$\lambda= -\ln\left(1-(1/\overline{Y})\right)$$
But my answer is
$$\lambda= \ln\left(1+(1/\overline{Y})\right)$$
Where am I getting it wrong.
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 $\hat{\lambda}$.