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Take $n$ non-negative dependent random variables $X_1,…, X_n$ with $P(X_i \leq t) = t, t \in [0,1]$, and consider $X = \min_i X_i$.

Show that $E[X] \leq 1/2$ and $E[X] \geq 1/(2n)$, the lower bound using the union bound to obtain $P(X \leq t)$ and applying the definition of expectation.

Applying directly the union bound to the variables would get something like $P(X >t) \leq \sum_i P(X_i > t) = n(1 - t)$, and $P(X \leq t) \geq 1 - n(1-t)$. Considering the definition of expectation:

$$E[X] \geq \int_0^1 tn dt = \frac{n}{2}$$

I believe I've made a mistake somewhere, as this lower bound is higher than the upper bound and looks nothing like the one presented. Is there maybe an additional inequality I am not using?

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The event $\left\{X\lt t\right\}$ is equal to $\bigcup_{i=1}^n\left\{X_i\lt t\right\}$ hence for all positive $ t$, $$\mathbb P\left\{X\lt t\right\}\leqslant \min\left\{nt,1\right\}$$ which implies that $$ \mathbb P\left\{X\geqslant t\right\}\geqslant 1-\min\left\{nt,1\right\}. $$ Integrate this over $(0,+\infty)$.

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  • $\begingroup$ Isn't that similar to the original solution but considering the opposite events? The expression $\min\{nt,1\}$ makes it difficult to calculate the integral, but I don't think this will be equal to $1/(2n)$. $\endgroup$
    – user550496
    Apr 9, 2018 at 10:22
  • $\begingroup$ It is better to work with the union of $\left\{X_i\lt t\right\}$ since the bound is tighter. The integral is perfectly computable: divide it into two pieces: $\int_0^{1/n}+\int_{1/n}^{\infty}$. $\endgroup$
    – Davide Giraudo
    Apr 9, 2018 at 10:25
  • $\begingroup$ You are absolutely right, thank you very much! $\endgroup$
    – user550496
    Apr 9, 2018 at 10:33

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