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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.

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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?

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  • $\begingroup$ Hmm, I'm not sure what you mean by conclude, since you gave the answer :D So, the probability density function of $Y$ is: $f(y)=\frac{1}{\sqrt{2\pi H_n}}e^{-\frac{y^2}{2H_n}}$, where $H_n$ is $n$-th harmonic number. $\endgroup$
    – A6EE
    Commented Jul 11, 2018 at 17:17
  • $\begingroup$ By "conclude" I mean put things together: since $Y$ has the same c.f. as that of a Gaussian variable with mean 0 and variance $\sigma_n^2$, then $Y$ is distributed as a Gaussian variable with mean 0 and variance $\sigma_n^2$. $\endgroup$
    – Clement C.
    Commented Jul 11, 2018 at 17:19
  • $\begingroup$ Yes, thank you :) $\endgroup$
    – A6EE
    Commented Jul 11, 2018 at 17:20
  • $\begingroup$ @A6EE You're welcome! $\endgroup$
    – Clement C.
    Commented Jul 11, 2018 at 17:22

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