what is $F_Y(y)$ based on $X_i$ I am given random variables $X_i$, where $i=1, 2, \ldots ,n$. They are all independent and have the distribution of $N(0,1)$. I also have a random variable $Y=X_1 + 2X_2 + \cdots + nX_n$ and now I am supposed to find the distribution of a random variable
$$Y^* = \frac{Y-E(Y)}{\sqrt{D(Y)}}$$
What I have already tried is this:
$F_y(Y)=P(Y<y)=P(X_1 + 2X_2 + \cdots + nX_n < y)$, now, since the sum of these random variables is less than $y$, that means that each of them is also less than $y$. Now, if I am correct, I get the following 
$P(X_1<y)\cdot P(2X_2 < y)\cdots P(nX_n<y)$ and this is where I get stuck. How do I continue from here and get the cumulative distribution function of each of these variables? Any hints are appreciated, thanks
 A: The key is that the sum of a series of independent normally distributed random variables is a normally distributed random variable whose mean is the sum of the series' means and variance the sum of the series' variances.
Also if $X_k\sim\mathcal N(0,1^2)$ then $kX_k\sim\mathcal N(0, k^2)$ 


*

*Generally: $Z\sim \mathcal N(\mu_Z,\sigma_Z^2)\implies aZ+b\sim\mathcal N(a\mu_Z+b, a^2\sigma^2_Z)$


So, therefore: $$\sum_{k=1}^n k X_k~\sim~\mathcal N\left(0, \sum_{k=1}^n k^2\right)$$
Take it from there.

So we have $Y\sim \mathcal N(\mathsf E(Y), \mathsf D(Y))$ for some parameters (specifically $0$ and $\sum_{k=1}^n k^2$ as above).
Given that then since $Y^*$ is a linear function of $Y$, namely $\frac{Y-\mathsf E(Y)}{\surd \mathsf D(Y)}$, therefore ...
A: $$P(X_1 + 2X_2 + \cdots + nX_n < y)$$ is definitely not equal to $$P(X_1<y)\cdot P(2X_2 < y)\cdots P(nX_n<y)$$ since the event $$Y<y$$ is not the same as $$\Big[ (X_1 <y)\ \&\ (2X_2<y)\ \&\ \cdots\ \&\ (nX_n<y)\Big].$$
Notice that
\begin{align}
X_1 & \sim N(0,1) \\
2X_2 & \sim N(0,4) \\
3X_3 & \sim N(0,9) \\
& \,\,\,\vdots \\
nX_n & \sim N(0,n^2) \\[5pt]
\text{so } X_1+ 2X_2 + \cdots+nX_n & \sim N(0,1+4+9+\cdots+n^2)
\end{align}
But you don't need that in order to answer the question: You just need to know that that sum is normally distributed, and so, since
\begin{align}
& \operatorname{E} \frac{Y - \operatorname{E}(Y)}{\sqrt{\operatorname{var}(Y)}} = 0 \\[10pt]
& \text{and } \operatorname{var} \frac{Y-\operatorname{E}(Y)}{\sqrt{\operatorname{var}(Y)}} = 1, \\[10pt]
&\text{we conclude that } \frac{Y-\operatorname{E}(Y)}{\sqrt{\operatorname{var}(Y)}} \sim N(0,1).
\end{align}
A: This is related to @Graham-Kemp answer.
$$ Y = X_1 + 2X_2 + \cdots + nX_n $$
We can determine the distribution of $Y$ by finding it's moment generating function (mgf) : $E[e^{tY}]$. 

The mgf of a normal distribution with parameters $\mu$ and $\sigma$
  is  $$ e^{\mu t + \frac{1}{2}(\sigma t)^2} $$

Since $X_{i}$'s are identical and independent, I may write
$$ \operatorname{mgf}(Y) = E[e^{tY}] = E[e^{tX_{1} + 2tX_{2} + \cdots + ntX_{n}}] = E[e^{tX_{1}}]E[e^{2tX_{2}}]E[e^{3tX_{3}}]\cdots E[e^{ntX_{n}}] $$
$$ \operatorname{mgf}(Y) = e^{\frac{1}{2}t^{2}+\frac{1}{2}(2t)^{2} + \cdots + \frac{1}{2}(nt)^2} = e^{\frac{1}{2}t^{2}(\sum_{i=1}^{n} i^{2})}$$
So I may conclude that $ Y $ is normally distributed with mean $\mu = 0$ and variance $Var = \sum_{i=1}^{n} i^{2}$. Actually, the information : $Y$ is normally distributed is sufficient to answer your question. This is because if $X$ is normally distributed (with any mean and variance) then $Z = \frac{X-\mu_{X}}{\sigma_{X}}$ is normal with mean $0$ and variance $1$. So you can conclude the distribution of $Y^{*} = \frac{Y - \mu_{Y}}{\sigma_{Y}}$. 
  Thanks.
