## fisher information comparison?

The aim is to estimate the probability of success $θ ∈ (0, 1)$ of a certain event. You have two options to sample:

a) $X_1, X_2, \ldots , X_n$ from $\operatorname{Bernoulli} (θ)$ distribution $f_{X_1} (x, \theta) = \theta^x(1 - \theta)^{1-x}, x = {0, 1}.;$

b) $Y_1, Y_2, \ldots , Y_n$ from a geometric distribution: $f_{Y_1}(y, \theta) = \theta (1 - \theta)^{y-1}, y = 1, 2, \ldots$

Which type of drawn sample will be more informative about $θ$? If the sample size $n$ is large enough, which sampling will you recommend and why?

I worked out, the fisher information for each to be $I_X(\theta) = \frac{n}{\theta(1 - \theta}$ and $I_Y(\theta) = \frac{n}{\theta^2 (1-\theta)}$. Since $\theta <1, I_X(\theta) < I_Y(\theta)$ . Is this how it is done? By comparing the fisher information. Also, I'm not sure how to determine this question, "If the sample size n is large enough, which sampling will you recommend and why?"