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Let $X_1, \ldots , X_n$ be a random sample from a $N(0,1)$ population. Define $Y_1=|\frac1n \sum_{i=1}^nX_i|$ and $Y_2=\frac1n\sum^n_{i=1}|X_i|$. Find a relationship between $E(Y_1)$ and $E(Y_2)$.

I have a feeling that I will need to use Jensen's Inequality here. Since $X_i \sim N(0,1)$, the linear combination $\sum_{i=1}^nX_i \sim N(0,n)$.

$E(Y_1)=\frac1n \cdot E(|\sum_{i=1}^nX_i|)$

$E(Y_2)=\frac1n \cdot E(\sum_{i=1}^n|X_i|)$

However, I am not too sure how to compute $E(Y_1)$ or $E(Y_2)$ from this step since there are absolute values. Do I have to use a cdf approach?

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    $\begingroup$ Can you find a relationship between $Y_1$ and $Y_2$ that holds pointwise? $\endgroup$
    – user257566
    Nov 16 '19 at 22:28
  • $\begingroup$ @user257566 I am not sure if I know what you mean $\endgroup$
    – Jen Snow
    Nov 16 '19 at 23:09
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    $\begingroup$ For example, for numbers $a$ and $b$, what can you say about $|a+b|$ compared to $|a| + |b|$? $\endgroup$
    – user257566
    Nov 16 '19 at 23:40
  • $\begingroup$ @user257566 $|a| + |b| \ge |a+b|$. So, $\Sigma^n_{i=1}|X_i| \ge |\Sigma_{i=1}^n X_i|$. However, I would like to show a rigorous proof of this and potentially calculate $E(Y_1)$ and $E(Y_2)$. $\endgroup$
    – Jen Snow
    Nov 17 '19 at 4:08
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    $\begingroup$ that's right. now, using that $Y_2$ is pointwise larger than $Y_1$ to (rigorously) prove your result just relies on a simple use of that bigger RVs have bigger means (see e.g. thm 1.6.1 of durrett probability, v5) $\endgroup$
    – user257566
    Nov 17 '19 at 18:05
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Assuming independent $X_i$, the mean $\frac{1}{n}\sum X_i$ is also normal, i.e. $N(0,1/n)$. Absolute value of it is Half-normal, which has mean $E[Y_1]=\frac{\sigma\sqrt{2}}{\sqrt{\pi}}=\sqrt{\frac{2}{n\pi}}$. For $Y_2$ we can find the expected value directly: $$E[Y_2]=\frac{1}{n}\sum_{i=1}^n E[|X_i|]=E[|X_i|]=\sqrt\frac{2}{\pi}$$

This means $\sqrt n E[Y_1]=E[Y_2]$. I think an equality is better than inequality.

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  • $\begingroup$ I understand how you got that $\frac{1}{n}\Sigma X_i \sim N(0,1/n)$. Is there a way to find $E(Y_1)$ without having to use the half-normal distribution? $\endgroup$
    – Jen Snow
    Nov 19 '19 at 0:24
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    $\begingroup$ Calculating $E[|X|]$ for a zero-mean normal RV is straightforward (let $x^2/2\sigma^2=u$): $$E[|X|]=2\int_0^\infty\frac{1}{\sqrt{2\pi}\sigma}xe^{-x^2/2\sigma^2}dx=2\frac{\sigma}{\sqrt{2\pi}}\int_0^\infty e^{-u}du=\sigma\sqrt{\frac{2}{\pi}}$$ $\endgroup$
    – gunes
    Nov 19 '19 at 6:25
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Answer:

Whatever the distribution of $X_1,...,X_n$,

$$\mathbb{E} Y_2 \geq \mathbb{E} Y_1.$$

Details:

For any $n$ numbers $X_1,..., X_n$ it is true that

$$ \sum_i |X_i| \geq |\sum_i X_i|$$

and dividing both sides by $n$:

$$ \frac{1}{n}\sum_i |X_i| \geq \frac{1}{n}|\sum_i X_i| = |\frac{1}{n}\sum_i X_i|.$$

Now, the key word is 'any', that means that if we take integrals of both sides of the above inequality, the integrand of the left hand side is uniformly higher than the integrand of the right hand side, i.e.

$$ \mathbb{E} Y_2 = \int \frac{1}{n}\sum_i |X_i| dX_1 ...dX_n \geq \int |\frac{1}{n}\sum_i X_i| dX_1...dX_n = \mathbb{E} Y_1.$$

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