I am stuck by the following problem. Could you please help me? Million thanks.
We have $c$ non-negative, i.i.d. random variables $X_1, X_2, ..., X_c >0$, where $\mathbb{E}[X_i]\geq\lambda>0$, and $X_i-\mathbb{E}[X_i]$ is sub-gaussian with variance $V_i\leq \beta^2$, $\forall i\in[c]= \{1,2,...,c\}$.
Now consider $Y=\min_{i\in[c]} X_i$, I want to prove a lower bound on $\mathbb{E}[Y]$, i.e., $\mathbb{E}[Y]\geq f(\lambda,c,\beta)>0$, where $f(\lambda,c,\beta)$ is a positive quantity that depends on $\lambda$, $c$ and $\beta$.
Below is what I have tried.
From the property of sub-gaussian random variables:
$P(X_i-\mathbb{E}[X_i]<-a)\leq e^{-\frac{a^2}{2\beta^2}}$, $\forall a>0$.
Therefore, for a fixed $i$:
$P(X_i\geq\lambda-a)\geq 1-e^{-\frac{a^2}{2\beta^2}}$.
Thus,
$P(\min_{i\in [c]}X_i>\lambda-a)\geq (1-e^{-\frac{a^2}{2\beta^2}})^c$.
So we have
$\mathbb{E}[Y]\geq\lambda_2=\int_{0}^{\lambda}(1-e^{-\frac{(\lambda-x)^2}{2\beta^2}})^c dx$.
I am wondering if it is possible to get a better result, or if it is possible to simplify this to get rid of the integration. Could you please kindly help me with this? Thanks a lot!