Let $X_1$, $X_2$, …, $X_n$ be independent random variables with normal distribution and:
$$E(X_i)=0\\Var(X_i)=1\\i=1,2, …, n$$
We define random variable $Y$ as:
$$Y=\frac{X_1}{1}+\frac{X_2}{\sqrt{2}}+\cdot\cdot\cdot+\frac{X_n}{\sqrt{n}}$$
The task is to find the distribution of $Y$.
Now, after several steps of finding characteristic functions for $X_i$:
$$\rho_{X_i}(t)=e^{-\frac{t^2}{2}}$$
and using some characteristic function properties, I found that the characteristic function of $Y$ is:
$$\rho_Y(t)=\prod\limits_{k=1}^{n}\rho_{X_k}\bigg(\frac{t}{\sqrt{k}}\bigg)$$
$$=\prod\limits_{k=1}^{n}e^{-\frac{t^2}{2k}}$$
$$=e^{-\frac{t^2}{2}\sum\limits_{k=1}^n\frac{1}{k}}$$
I'm not sure how I can transform that expression to the general form:
$$\rho(t)=\int\limits_{-\infty}^{\infty}e^{itx}f(x)dx$$
from where I can extract the density function and find distribution.
Best Answer
Set $\sigma_n = \sqrt{\sum_{k=1}^n \frac{1}{k}}$ (i.e., the square root of $n$ harmonic number). Then, as you have shown, $$ \forall t\in\mathbb{R}, \qquad\rho_Y(t) = \mathbb{E}[e^{itY}] = e^{-\sigma_n^2\frac{t^2}{2}} \tag{1} $$ Looking at the characteristic of a Gaussian r.v. (which you used to find $\rho_{X_k}$), you can notice that this is the characteristic function of a $\mathcal{N}(0,\sigma_n)$ random variable (Gaussian with mean $0$ and standard deviation $\sigma_n$). Now, since the characteristic function characterizes the distribution, can you conclude?