Suppose that $Y_1,Y_2,...,Y_n$ are i.i.d. with distribution $exp(\lambda)$, and we define another set of random variables $Z_1,Z_2,...,Z_n$ where $Z_j=\delta \lfloor \frac{Y_j}{\delta} \rfloor$.
I have successfully found the likelihood contribution from $z_j$ to be $$(1-e^{-\lambda \delta})(e^{-\lambda z_j})$$ and that the expected information for $\lambda$ is $$\frac{n\delta ^2 e^{-\lambda \delta}}{(1-e^{{-\lambda \delta}})^{-2}}$$
Now suppose that $\lambda = 1$. Show that the loss of information when data (from the $exp(1)$ distribution) are rounded down to the nearest integer, is less than 10%.
So, from what I gather, since we are now rounding down to the nearest integer, we have simply $Z_j=\lfloor Y_j \rfloor$, i.e. $\delta = 1$.
Is there some sort of technical definition for "loss of information" that I am not aware of? Or is the question just asking for the expected size of the truncated part as a ratio of the actual value, i.e. to calculate $\Bbb E (\frac{Y_j - Z_j}{Y_j})$?
It seems that $\Bbb E (\frac{Y_j - Z_j}{Y_j}) >> 0.1$ though...
I think here, the "loss of information" amounts to calculating the difference between the Fisher information for the exponential distribution, and the Fisher information from rounding, with $\lambda = 1$ (given), and $\delta = 1$, as you said.
In other words, we want $$ \frac{n}{\lambda^2} - \frac{n\delta^2e^{-\lambda\delta}}{(1-e^{-\lambda\delta})^2} $$ with $\lambda = \delta = 1$, so $$ n\bigg( 1 - \frac{e^{-1}}{(1 - e^{-1})^2} \bigg) $$ and evaluating what is in the parentheses gives you approximately 0.079.