Finding order statistic of $Y_{\min}$

Question

The question below is asking me to find the $Y_min$ such that it is smaller than $0.2$ and has a probability bigger than $0.9$, if I'm understanding it correctly.

Let $Y_1, . . . , Y_n$ be iid random variables with exponential pdf $f_Y (y) = e^{−y}, y ≥ 0$. What is the smallest $n$ for which $P(Y_{\min} < 0.2) > 0.9$?

Therefore, I followed this formula:

where $F_Y(y)= -e^{-y}$ and I got:

$$\int^{0.2}_0 n[1+e^{-y}]^{n-1} dy = 0.9$$ But apparently, I was supposed to follow formula instead? I don't understand why since it's used to find the max order statistics

Answer 1

I assume your teachers are not crazy and $F$ is the CDF, not the tail function.

In that case you have a typo $F_Y(y) = 1 -e^{-y}$.

In that case again, you have the naming of the two formulas confused. The first formula is for the max, the second is for the min.

Answer 2

I would not have used integration and instead would have expected to use the fact that the minimum exceeds a value if all the individual cases exceed that value:

$\mathbb P(Y_{\min}\le y)=F_{Y_{\min}}(y) $ $= 1-\mathbb P(Y_{\min}>y) $ $= 1-(1-F_Y(y))^n $ $= 1-\left(e^{-y}\right)^n $ $= 1-e^{-ny}$

So the answer here with $y=0.2$ and $n=10$ is $1-e^{-2}$