Let $X_n$ be a normally distributed random variable ($N(0,\sigma_n^2)$), $Y_n$ is gamma distributed ($\Gamma(p_n,a_n)$). $X_n$ and $Y_n$ are independent. Let $Z_n := X_n \sqrt Y_n + 2Y_n$, $n = 1,2,\ldots$ I want to find the characteristic function of $Z_n$ and find such $a_n, p_n, \sigma_n^2$ so that $Z_n$ converges to $N(0,1)$ in distribution.

$$\phi_{Z_n}(t) = \mathbb E (\mathbb E (e^{it(X_n \sqrt Y_n + 2Y_n)}) \mid Y_n)$$

Because of the independence of $X_n$ and $Y_n$ we obtain

$$\mathbb E(e^{it(X_n \sqrt Y_n + 2Y_n)} \mid Y_n = y) = e^{it2y} \mathbb E(e^{it\sqrt y X_n}) = e^{it2y} e^{-\frac{t^2 \sigma_n^2}{2}y}$$


$$\phi_{Z_n}(t) = \mathbb E\left(\exp\left(itY_n\left(2-\frac{t^2\sigma_n^2}{2it}\right)\right)\right)$$

Using the characteristic function of gamma distribution I got:

$$\phi_{Z_n}(t) = \Biggl(1-\frac{2it - (\frac{t^2 \sigma_n^2}{2})}{a_n} \Biggr)^{-p_n}$$

That's the place where I am stuck. My calculations may be wrong, I can't find such sequences so that I get the characteristic function of standard normal distribution.

  • $\begingroup$ Why do you expect it to converge to a normal distribution? The C.L.T. does not apply. $\endgroup$ – herb steinberg Aug 25 '18 at 19:33
  • $\begingroup$ @herbsteinberg That's an exercise, so I guess it does converge to standard normal distribution with proper parameters. But I can't find proper sequences. For example, if $\sigma_n^2 = 1, a_n = p_n = n$, then $Z_n$ converges to $N(2,1)$ $\endgroup$ – treskov Aug 25 '18 at 19:42
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    $\begingroup$ $E(Z_n)=2E(Y_n)$, so unless you can make $E(Y_n)\to 0$, you will never get a normal distribution with mean $0$. $\endgroup$ – herb steinberg Aug 25 '18 at 20:11
  • $\begingroup$ @herbsteinberg Can't we make $E(Y_n) \rightarrow 0$? It requires $p_n \over a_n$ converging to $0$. $\endgroup$ – treskov Aug 25 '18 at 20:28
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    $\begingroup$ Let $H=2it-\frac{t^2\sigma_n^2}{2}$. Then $\phi_{Z_n}(t)=e^{-p_nlog(1-\frac{H}{a_n})}\approx e^{\frac{p_n}{a_n}H}$. When $\frac{p_n}{a_n}\to 0$, then $\phi _{Z_n}(t)\to 1$. $\endgroup$ – herb steinberg Aug 25 '18 at 20:53

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