1
$\begingroup$

We have that $X_1,X_2,...$ are iid exponential variables having mean 1. $Y_1,Y_2,...$ are iid Poisson random variables having mean 1. $Z_1, Z_2,...$ are iid standard normal random variables. $X_i, Y_i, Z_i$ are all independent.

$$V_n = \frac{\sum (X_i-Y_i+Z_i)}{\sum Y_i^2}$$

I am trying to find the pdf of the limiting distribution of the sequence $V_1,V_2,...$

We can use the CLT and let $T_i = X_i-Y_i+Z_i$ and therefore we have $$\frac{\sum T_i}{\sqrt{n}\sigma_T}$$ where $\sigma_T$ is the standard deviation of the $T_i$. Additionally, $V_n$ can be expressed as a product of two or three factors, with one factor being easily dealt with using the weak law of large numbers, another factor being easily dealt with using the CLT, and the third may be a constant.

But I am unsure where to take it from here to get the limiting distribution

$\endgroup$
0
$\begingroup$

Hint: $\frac{\sum T_i}{\sqrt{n}\sigma_T}$ converges in distribution to a standard Normal random variable by the CLT, and $\frac{\sum_i Y_i^2}{n}$ converges in probability to $2$ by the WLLN. Then apply Slutsky's theorem.

$\endgroup$
  • $\begingroup$ how does it converge in probability to 2 if the mean is 1? $\endgroup$ – statrat403 Feb 27 '18 at 17:31
  • $\begingroup$ @chelebele403 it converges to the second raw moment which is the variance plus the square of the mean. $\endgroup$ – Taylor Feb 27 '18 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.