2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Jul 11 '18 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. Jul 11 '18 at 17:19
  • $\begingroup$ Yes, thank you :) $\endgroup$ – A6EE Jul 11 '18 at 17:20
  • $\begingroup$ @A6EE You're welcome! $\endgroup$ – Clement C. Jul 11 '18 at 17:22

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.