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I have a random Variable $X$ is $ SN(\lambda)$ and is pdf is given by: $f(x)=2\phi(x)\Phi(\lambda x)$.

The model of the variable X is given by:$X=\frac{1}{\sqrt{1+\lambda^2}}Z_1+\frac{\lambda}{\sqrt{1+\lambda^2}}|Z_2|$ where $Z_1$ and $Z_2$ have $N(0,1)$ and are independent.

What I need to do is to find the moment estimator for $\lambda$.

I've been reading about the method of moments but don't really understand what I'm supposed to do for this exercise. If you could show how to solve it or at least the first steps, that would be a huge help.

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Steps for method of moments:

  1. Get the theoretical moments: In your case, $\mathrm{E}(X) = \frac{\lambda}{\sqrt{1+\lambda^2}}\mathrm{E}(|Z_2|)$

  2. Let sample moments = theoretical moment: $\bar {X} = \frac{\hat \lambda}{\sqrt{1+\hat \lambda^2}}\mathrm{E}(|Z_2|)$

  3. Resolve the parameters:

So you need to find $\mathrm{E}(|Z_2|)$ and the solution of $\hat \lambda$.

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  • $\begingroup$ Why is the expected value of $X$ just $\frac{\lambda}{\sqrt{1+\lambda^2}}\mathrm{E}(|Z_2|)$? Shouldn't it be $\frac{1}{\sqrt{1+\lambda^2}}\mathrm{E}(|Z_1|)+\frac{\lambda}{\sqrt{1+\lambda^2}}\mathrm{E}(|Z_2|)$? $\endgroup$ – dmalka May 30 '17 at 2:09
  • $\begingroup$ Your one condition is Z1 following standard normal, so E(Z1) =0. Or you missed | | around Z1 in your question? If so, your equation in comment is correct. $\endgroup$ – user158565 May 30 '17 at 2:24

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