3
$\begingroup$

So I have stumbled upon this problem. Let $X_1, \dots, X_n \sim N(\mu, \sigma^2)$ be iid. Define: $$S = \frac{1}{n}\sum_{i=1}^n I[X_i > a]$$ $$T = I[\frac{1}{n}\sum_{i=1}^n X_i > a]$$ $a > 0$. Using Jensen's Inequality prove: $$E(S) > E(T)$$ Now I only manage to prove it by solving the expected values without Jensen's Inequality. Where I get: $$E(S) = 1 - \Phi\left(\frac{a-\mu}{\sigma}\right)$$ And $$E(T) = 1 - \Phi\left(\frac{a-\mu}{\sigma}\sqrt n\right)$$ Which proves the inequality. Where $\Phi$ is the standard normal cdf. However this is just by using $E(f(X)) = \int_{-\infty}^{\infty} f(x)p(x) dx$. $p(x)$ is the pdf of $X$.

I struggle seeing why one can apply Jensen on $I(X > a)$ as it is non-convex.

Edit: After some thinking I do not belive this is possible, but feel free to prove me wrong.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think the proof is not valid.

let $n>1$,

if $a-\mu >0$

$$ \left(\frac{a-\mu}{\sigma} \right) < \left(\frac{a-\mu}{\sigma} \right)\sqrt{n}$$

so

$$\Phi \left(\frac{a-\mu}{\sigma} \right) <\Phi \left(\frac{a-\mu}{\sigma} \sqrt{n}\right)$$

But $a-\mu <0$

$$ \left(\frac{a-\mu}{\sigma} \right) > \left(\frac{a-\mu}{\sigma} \right)\sqrt{n}$$

so

$$\Phi \left(\frac{a-\mu}{\sigma} \right) >\Phi \left(\frac{a-\mu}{\sigma} \sqrt{n}\right)$$

So the bigger value of

$$E(S) = 1 - \Phi\left(\frac{a-\mu}{\sigma}\right)$$ And $$E(T) = 1 - \Phi\left(\frac{a-\mu}{\sigma}\sqrt n\right)$$

depend on $a<\mu$ or $a>\mu$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .