[Math] Find a the density function for Y(1) = min(Y1, Y2, . . . , Yn).

probability distributions

Suppose that the length of time Y it takes a worker to complete a certain task has the probability density function given by
$f(y)=\begin{cases} e^{-(y-\theta)} &, y>\theta\\ 0 & ,elsewhere \end{cases}$
where θ is a positive constant that represents the minimum time until task completion. Let
$Y_1, Y_2, . . . , Y_n$ denote a random sample of completion times from this distribution. Find

a the density function for Y(1) = min(Y1, Y2, . . . , Yn).
b E(Y(1)).

Could anyone get me started on this since the random sample is not independent?

Best Answer

Hints:

  • The sample times are presumably independent.
  • Find the probability $Y_i \le x$ for a particular $i$
  • Find the probability $Y_i \gt x$ for a particular $i$
  • Find the probability $Y_i \gt x$ for all $i$
  • Find the probability $\min_i(Y_i) \gt x$
  • Find the probability $\min_i(Y_i) \le x$
  • Find the density function for $\min_i(Y_i)$
  • Find the expectation of $\min_i(Y_i)$
Related Question