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# Metrics Final 2020 Q1
## Question

Two sequences of independent random variables $\{X_n\}_{n=1}^\infty$ and $\{Y_n\}_{n=1}^\infty$ are such that

$$
\sqrt{n} X_n \to_d N(1,1)
$$

and

$$
\sqrt{n} Y_n \to_d N(10,10)
$$

We construct a new sequence of random variables

$$
Z_n= \begin{cases}
X_n & \text{ if $n$ is odd},\\
Y_n & \text{ if $n$ is even},
\end{cases}
$$

1. Does $Z_n$ converge in probability? Prove or disprove.
2. Does $Z_n$ converge almost surely? Prove or disprove.
3. Does $\sqrt{n} Z_n$ converge in distribution? Prove or disprove.

## Answer
### Converge in Probability

Define $W_n = \sqrt{n} X_n$. Let $F_{W_n}$ and $F_{X_n}$ be the distribution function of $W_n$ and $X_n$, respectively. We have

$$
F_{X_n}(x) = P(X_n \le x) = P(W_n \le x \sqrt{n} ) = F_{W_n}( x \sqrt{n} ).
$$

Let $F$ be the distribution function of $N(1,1)$. Since $W_n \to_d N(1,1)$, for any $a$ and $\varepsilon$, we can find large enough $n$ such that

$$
|| F_{W_n}(a) - F_{W_n}(-a) | - | F(a) - F(-a) || \le \varepsilon.
$$

For any $\varepsilon>0$, $P(X_n > \varepsilon) + P(X_n <\varepsilon)$ is,

$$
1- F_{W_n}(\varepsilon \sqrt{n}) + F_{W_n}(-\varepsilon \sqrt{n}) = 1 - \left (F_{W_n}(\varepsilon \sqrt{n}) - F_{W_n}(-\varepsilon \sqrt{n})\right).
$$

Hence, with large enough $n$, $P(X_n > \varepsilon) + P(X_n <\varepsilon)$ can be arbitrarily small. That is, the limit distribution of $X_n$ is a degenerate distribution of $0$. Since converge in distribution to a constant implies converge in probability, $X_n \to_p 0$.

Similary, $Y_n \to_p 0$. Hence, $Z_n \to_p 0$.

### Converge almost surely

Let $(\Omega, \mathcal{F}, P)$ be the underlying probability space. Suppose $\omega \in \Omega$ such that $\lim_{n \to \infty} X_n(\omega) \neq 0$. In other words,

$$
\lim_{n \to \infty} \frac{W_n(\omega)}{n^{0.5}} \neq 0 \implies \limsup_{n \to \infty} W_n(\omega)> n^{0.25}.
$$

However, $\lim_{n \to \infty} P(W_n > n^{0.25})=0$, hence,

$$
P\left( \{\omega: \limsup_{n \to \infty} W_n(\omega) > n^{0.25}\}\right) =0.
$$

Therefore, $X_n \to_{a.s} 0$ and $Y_n \to_{a.s} 0$. As a result, $Z_n \to_{a.s} 0$.