The life of a repairing device is $Exp(1/a)$-distributed. Peter wishes to use it on $n$ different, independent, $Exp(1/na)$-distributed occasions. Compute the probability $P_n$ that this is possible.
This question has been posted before and got a good answer. My question is why my different approach to this question gives me the wrong answer:
Let $X_1,X_2,..X_n$ denote the $n$ independent times he uses the device, where $X_i \in Exp(1/na). $Let $U_n$ be the time it takes for Peter to use the device $n$ times. Let $L$ be the life time of the device. We wish to compute $P_n=P(U_n<L)$.
Now, $U_n=max(X_1,X_2,....X_n)$, so we get that $F_{U_n}(t)=(F_X(t))^n$, so: $$P_n=P(U_n<L)=\int_0^\infty P(U_n<L|L=t)\cdot f_L(t)dt=\int_0^\infty P(U_n<t)\cdot f_L(t)dt=$$ $$\int_0^\infty (F_X(t))^n\cdot f_L(t)dt$$
But this gives me the wrong answer. The correct one is $P_n=(\frac{n}{n+1})^n$.
Where is my thinking wrong? Have i missed any dependece property?
You made a conceptual mistake right in the beginning. Your definition of $U_n$$$U_n=\mathrm max(X_1,\ldots,X_n)$$ does not describe the time it takes Peter to use the repairing device over all occasions but gives you the longest time for one occasion. You would have to sum up the repairing times to get the overall time.
Conceptually I'd suggest to view this problem in a different way: Let $X_i$ be the repairing occasions and $L_i$ be the life of the device, then you actually want to compute $P[\cap_{i=1}^n (X_i< L_i)]$ and then you can use the fact that all source of randomness are independent.