$X_1$,$X_2$ ~ $P(1)$, $Y$ = Min{$X_1$,$X_2$} $\mathbb\quad P(Y=1)$ =$?$
I tried this in two different ways and I am getting a different answer in both:-
- Firstly I made PDF of $X_{(1)}$ = $n(1-F(x))^{n-1}f(x)$ = $$\frac{2e^{-1}}{x!}(1-\sum_{x=0}^x \frac{e^{-1}}{x!}) $$ Then after Putting the value $\mathbb1$, I get, $$\frac{2e-4}{e^2}$$
- $\mathbb\quad P(Y=1)$ = $\mathbb\quad P(X_1=1)P(X_2>=1)+P(X_2=1)P(X_1>=1)-P(X_1=1)P(X_2=1)$ = $$\frac{2e-3}{e^2}$$ Second answer is correct but I don't understand what I have done wrong in first part. Is it that this PDF of order stats is good in continuous cases as what I think the PDF didn't include the case when both $X_1$,$X_2$ are equal to 1. Kindly help.
Your formula in the first part is wrong.
$P(X_{(1)}\leq x)=1-P(X_1>x)^{2}=1-(1-F(x))^{2}$. So $P(X_{(1)}= 1)=(1-F(0))^{2}-(1-F(1))^{2}=(1-e^{-1})^{2}-(1-2e^{-1})^{2}=\frac {2e-3}{e^{2}}$.