$\mathbb{E}[\sqrt{1+X_1^2+...X_n^2}]=\sigma \sqrt n$

Question

Let $X_1,X_2,...$ be i.i.d with $\mathbb{E}[X_1]=0$ and $\mathbb{E}[X_1^2]=\sigma^2$. Prove $$\mathbb{E}[\sqrt{1+X_1^2+...X_n^2}]=\sigma \sqrt n$$

It might not be true but I feel like it is (or at least at the limit it is).
I needed to solve a problem and if I could show this I would finish, but realized I have no idea how to show it and I'm not even sure if its true.

The original problem is

Prove $\frac{\sqrt{1+X_1^2+...X_n^2}}{\sqrt n}\overset{d}{\rightarrow}\sigma$ or $\frac{\sqrt{1+X_1^2+...X_n^2}}{\sqrt n}\overset{p}{\rightarrow}\sigma$

I'm taking a course in probability theory.

Thank you,

Answer 1

The equality you suggest may not be true for example if $X_i$ takes the values $-1$ and $1$ with probability $1/2$.

However, for your original problem, you can use the fact that if $(Y_n)_{n\geqslant 1}$ is a sequence of non-negative random variables which converges in probability to some $y$, then $\left(\sqrt{Y_n}\right)_{n\geqslant 1}$ converges in probability to $\sqrt y$ (this is a special case of this). Use this with $Y_n=\left(1+\sum_{i=1}^nX_i^2\right)/n$ and a well-known theorem.

Answer 2

The Markov inequality might be used in conjunction. Let's write $\lim_{n \rightarrow \infty}\mathbb{P}(\left|\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}-\sigma\right|\geq\epsilon)\leq\lim_{n \rightarrow \infty}\frac{\mathbb{E}\left(\left|\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}-\sigma\right|\right)}{\epsilon}$ then iff $\lim_{n \rightarrow \infty}\mathbb{E}\left(\left|\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}-\sigma\right|\right)=0$ we will have $\lim_{n \rightarrow \infty}\mathbb{P}\left(\left|\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}-\sigma\right|\geq\epsilon\right)=0$ which is convergence in probability. Therefore, the stronger condition $\lim_{n \rightarrow \infty}\mathbb{E}\left(\left|\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}-\sigma\right|\right)=0$ implies $\lim_{n \rightarrow \infty}\mathbb{E}\left(\sqrt{\frac{1+x_{1}^{2}+\dots x_{n}^{2}}{n}}\right)=\sigma$ using the fact that $\mathbb{E}\left(|X_{n}|\right)\geq \mathbb{E}\left(X_{n}\right).$