# Is it true $(X_n, Y_n) \rightarrow N( (\mu_1, \mu_2), diag(\sigma^2_1, \sigma^2_2))$ when $X_n$, $Y_n$ each goes to normal independently.

I have $$X_n \overset{d}{\rightarrow} N(\mu_1, \sigma^2_1)$$, and $$Y_n \overset{d}{\rightarrow} N(\mu_2, \sigma^2_2)$$, and $$X_n$$ and $$Y_n$$ are independent, for $$n \in \mathbb{N}$$.

(Note that it does not imply $$X_i$$ and $$Y_j$$ are independent where $$i \neq j$$)

Can we say as follows, $$(X_n, Y_n) \overset{d}{\rightarrow} N_2\left((\mu_1, \mu_2), \begin{bmatrix} \sigma^2_1 & 0 \\ 0 & \sigma^2_2 \end{bmatrix}\right)$$?

My intuition is strongly positive about this, but I need a proof for this.

• I have a strong feeling that characteristic functions will do the work. Commented Sep 29, 2018 at 14:10

## 1 Answer

Let $$(U,V)\sim N((\mu_1,\mu_2),\operatorname{diag}(\sigma_1^2,\sigma_2^2))$$. We claim $$(X_n,Y_n)\xrightarrow{d}(U,V)$$. Indeed, by Levy's theorem it suffices to check characteristic function \begin{align*} \varphi_{(X_n,Y_n)}(s,t)&= \mathbb{E}[e^{i(s,t)\cdot(X_n,Y_n)}]\\ &=\mathbb{E}[e^{isX_n}]\mathbb{E}[e^{itY_n}]\\ &\to\mathbb{E}[e^{isU}]\mathbb{E}[e^{itV}]\\ &=\mathbb{E}[e^{i(s,t)\cdot(U,V)}]\\ &=\varphi_{(U,V)}(s,t) \end{align*}