Convergence of stochastic process

Question

I'm stucked with this exercise. Hope someone can help me

$X_t = 3+a_t$ with $a_t \sim (0,\sigma^2)$. $a_t$ is iid.

$Y_n=\frac{1}{\sqrt{n+2}} \sum_{t=1}^{n} X_t $

Does $Y_n$ converge in distribution? What distribution?

Answer 1

You have

\begin{align} Y_n &= \frac{1}{\sqrt{n+2}}\sum_{i=1}^nX_i \\ &= \frac{1}{\sqrt{n+2}} \sum_{i=1}^n(3 + a_i) \\ &= \frac{3n}{\sqrt{n+2}} + \frac{1}{\sqrt{n+2}}\sum_{i=1}^n a_i \\ \end{align}

Because $a_i$ are iid and follows a normal distribution $N(0,\sigma^2)$, then $\sum_{i=1}^n a_i$ follow the normal distribution $N(0,n\sigma^2)$. Hence, $Y_n$ follows the normal distribution $$Y_n = N(\frac{3n}{\sqrt{n+2}} ,\frac{n}{n+2}\sigma^2 )$$ with mean of $\frac{3n}{\sqrt{n+2}}$ and variance $\frac{n}{n+2}\sigma^2$.

When $n \rightarrow +\infty$, $Y_n$ still has a normal distribution of variance $\sigma^2$ but its mean tends to $+\infty$.