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For all integer $N>1$, I am trying to show that for a gaussian (or even better any type) cumulative distribution function $F(\theta;\mu,\sigma)$ ($\mu$ and $\sigma$ are the mean and standard deviation):

\begin{equation} 1-F(\theta;\mu,\sigma/\sqrt{N})>(1-F(\theta;\mu,\sigma))^N \end{equation} where $\theta>\mu$, which means that \begin{equation} 1-F(\theta;\mu,\sigma/\sqrt{N})<1-F(\theta;\mu,\sigma) \end{equation} I am quite sure that the first equation is true from simulations but need some kind of proof one way or another (or none if my simulations are wrong).

Sam
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Suppose $X_1,\ldots,X_N$ are independent and identically distributed normal RVs with mean $\mu$ and std dev $\sigma$. The LHS is $P(\frac{X_1+\cdots+X_N}{N} \geq \theta)$ (why?), and the RHS is $P(X_1 \geq \theta)^N$.

The event $X_1 \geq \theta,\,\ldots,\,X_N \geq \theta$ implies $\frac{X_1+\cdots+X_N}{N} \geq \theta$, and therefore, $$P\left(\frac{X_1+\cdots+X_N}{N} \geq \theta\right) \geq P(X_1 \geq \theta,\,\ldots,\,X_N \geq \theta) = P(X_1\geq\theta)^N.$$

It is not difficult to see that the inequality is strict when $N \geq 2$. This stems from the fact that the set of $x_1,\ldots,x_n$ with $\frac{x_1+\cdots+x_N}{N} \geq \theta,\,\exists i x_i < \theta$ has positive probability. Then, one has $$P\left(\frac{X_1+\cdots+X_N}{N} \geq \theta\right) = P(X_1 \geq \theta,\,\ldots,\,X_N \geq \theta) + P(\frac{x_1+\cdots+x_N}{N} \geq \theta,\,\exists i x_i < \theta) > P(X_1 \geq \theta,\,\ldots,\,X_N \geq \theta),$$ followed by the same arguments above.

Lord Soth
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