Let $X_1,X_2,X_3,…$ be a sequence of i.i.d. $N(\mu,1)$ random variables. Then, find $$\lim_{n\to \infty} \frac{\sqrt{\pi}}{2n}\sum_{i=1}^{n}E(|X_i-\mu|).$$
My thoughts:
I don't have any rigorous way to start, but just some thoughts. First I thought about writing out the expectations and $E(|X_i-\mu|)=\int(x_i-\mu)f_x(x_i)dx$ where $f_x(x)$ is the normal pdf. But I don't know is this approach helpful?
Next I tried to use some inequality. First thing came to my mind is to use Markov's inequality $P\{X\geq a\}\leq\frac{E(X)}{a}$, but it didn't help me to progress much further.
How can I solve this? So any ideas or help please. Thanks.
$X_i-\mu$ is a standard normal random variable, hence $$ \mathbb{E}[|X_i-\mu|]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}$$ Therefore the sum is equal to $$\frac{\sqrt{\pi}}{2n}\cdot n\sqrt{\frac{2}{\pi}}=\frac{1}{\sqrt{2}}$$ for all $n$.
Even without computing, we can see that the sum is independent of $n$ because the random variables are i.i.d.