Can we have $ \frac{1}{2} X_n^2+\frac{1}{2} Y_n^2\to \frac{1}{2}\chi_1^2+\frac{1}{2}\chi_1^2 $? Let $X_n$ and $Y_n$ be two iid sequences of random variables with $N(0,1)$. Then $X_n^2\to \chi_1^2$ in distribution and $Y_n^2\to \chi_1^2$, can we have
$$
\frac{1}{2} X_n^2+\frac{1}{2} Y_n^2\to \frac{1}{2}\chi_1^2+\frac{1}{2}\chi_1^2
$$
in distribution? Or it should be $\frac{1}{2} \chi_2^2$?
 A: First, note that if for all $n$ both $X_n$ and $Y_n$ are $\mathcal N(0,1)$ distributed, then their squares are also both $\chi_1^2$ distributed. Therefore an equal (in distribution) sign would be more appropriate than a limit sign.
Maybe what you had in mind instead was that $(X_n)$ and $(Y_n)$ were two independent sequences both converging in distribution to $\mathcal N(0,1)$ ? In that case it indeed follows from the continuous mapping theorem that $X_n^2 \to^d\chi_1^2$ and $Y_n^2 \to^d\tilde\chi_1^2$ ($\chi_1^2 $ and $\tilde\chi_1^2 $ are independent).
Regardless, regarding your question, you can show (e.g. using characteristic functions) that for any two independent sequences of r.v. $(A_n)$ and $(B_n)$ with $A_n\to^dA$ and $B_n\to^dB$, it holds that $A_n + B_n \to^d A +B$. Therefore it is true that
$$ \frac{1}{2} X_n^2+\frac{1}{2} Y_n^2\to^d \frac{1}{2}\chi_1^2+\frac{1}{2}\tilde\chi_1^2 = \frac 1 2\left(\chi_1^2 + \tilde\chi_1^2\right) $$
But note that the sum of two independent $\chi^2$ variables with one degree of freedom is also a $\chi^2$ with two degrees of freedom, therefore
$$ \frac 1 2\left(\chi_1^2 + \tilde\chi_1^2\right) =^d \frac 1 2\chi_2^2$$
So the two formulations are equivalent.

Bonus : You can also notice that $X_n^2 + Y_n^2\to^d Z^2 + \tilde Z^2$ where $Z$ and $\tilde Z$ are independent $\mathcal N(0,1)$ so
$$ \frac{1}{2} X_n^2+\frac{1}{2} Y_n^2\to^d \frac 1 2 \left(Z^2 + \tilde Z^2\right) =^d \frac 1 2 \chi_2^2 $$
Which is once again the same thing.
