# Sum of indicators and application of Jensen's inequality

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.

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 $$a>\mu$$.