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Suppose that $Z \sim \mathrm{Bern}(p)$ and $Y_1,Y_2 \sim \mathcal{N}(0,1)$ be Gaussian random variables and let all of them be independent. Suppose I am given samples from the following random variable $X$ defined as $$ X= Z + \frac{Y_1}{ Y_2}. $$ Notice that $X$ is a Cauchy centred at $1$ if $Z=1$ and $0$ if $Z=0$. So my question is whether we can infer the probability $p$ from samples $X_1,\ldots,X_n$? I am wondering if there is a solution to this problem. Of course, one could use EM to solve this but I want a consistent estimator for $p$.

I found on the Wikipedia article on Cauchy distributions https://en.wikipedia.org/wiki/Cauchy_distribution#cite_note-rothenberg-17 that the parameter estimation from Cauchy samples is usually done by taking the central mean of order statistics ($0.24$ fraction, to be precise). But the order statistics for the mixtures can be tricky and no closed form expressions exist.

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  • $\begingroup$ As the Cauchy distr. is symmetric, you could use the sum of the signs of $X$ as an estimator of $p$. So $p = \sum_j I(X_j>0) / n$. $\endgroup$
    – Semoi
    Jul 15, 2017 at 20:02
  • $\begingroup$ I don't see a reason why that would estimate $p$. Have you taken into consideration the Cauchy centered at $1$? $\endgroup$ Jul 15, 2017 at 20:51
  • $\begingroup$ @Semoi: But your idea is very close. The cdf of $X$ is given by $F_X(x)=pF(x-1)+(1-p)F(x)$ where $F$ is the cdf of Cauchy. Using this we can compute $F_X(0.5)$ and from this we can solve for $p$. $\endgroup$ Jul 15, 2017 at 21:13
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    $\begingroup$ What does the phrase "central mean" mean? I've never heard it before $\endgroup$
    – Glen_b
    Jul 16, 2017 at 6:33
  • $\begingroup$ @Glen_b: If we have $2n+1$ samples and they are ordered into $X_{(-n)},\ldots,X_{(0)},\ldots,X_{(n)}$, the central mean refers to the quantity $\frac{\sum_{i=-k}^k X_{(i)}}{2k+1}$. $\endgroup$ Jul 16, 2017 at 18:36

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