1

Given a random sample $X_1, X_2, ..., X_n$ where each $X_i$ has pdf:

$$ f(x; \theta) = 3 \theta^3 x^{-4} $$

and $0 \lt \theta \le x \le \infty$. Show that the distribution function $F$ for $min_{1 \le i \le n}{X_i}$ is:

$$ F(x) = 1 - (\theta/x)^{3n} $$

My calculations:

$$ F(x) = \int_{-\infty}^x{f(u)du} = \int_\theta^x{3 \theta^3 u^{-4}du} = 3 \theta^3 -\frac{1}{3}(x^{-3} - \theta^{-3}) = 1 - (\theta/x)^3 $$

I do not get how $F(x) = 1 - (\theta/x)^{3n}$. Where does the $n$ come from?

s5s
  • 685
  • It appears to me that you are calculating the cdf of X, not the distribution of the min... – Lucas Morin Jan 15 '19 at 22:03
  • @Lcrmorin I am not sure I understand what the distribution of the mean is. That is, isn't the mean just one of the variables? How would it be different from F(x)? – s5s Jan 15 '19 at 22:20
  • 2
    The min as in "minimum". Min is one of the variable yes, but smaller than the others. So you have to take all the n variables into account in your calculations. Throw some dices, the minimum is more likely to be 1 or 2. – Lucas Morin Jan 15 '19 at 22:27
  • Many close variations of this question have been answered here: see this search. – whuber Jan 15 '19 at 22:56

0 Answers0