Related to order statistics?

Question

say there are $N$ independent and identically distributed random channels i.e., $Z_i \in \{ Z_1, Z_2,..., Z_i\}$ with PDFs $f_{Z_i}(z)$ ordered in ascending order. Then using order statistics the PDF of $Z_1 = \text{min}_i(Z_i)$ is given by

$f_{Z_1}(z) = N (1-F_{Z_i}(z))^{N-1}f_{Z_i}(z)$ ---(1)

My query is will equation (1) will be valid if the random variables are not independent and not identically distributed.

Any help in this regard will be highly appreciated.

Answer 1

Using the assumptions we can develop the equation you provided.

Let $T_1 = \min_{i}{Z_i}$, and let's find the PDF of $T_1$, by first deriving the CDF of $T_1$: $$P(T_1 \leq t) = P(\min_{i}{Z_i} \leq t) = 1-P(\min_{i}{Z_i} >t) = 1-P(Z_1>t,...,Z_N>t)$$ Here we use the fact that the random variables are independent to get: $$ = 1-P(Z_1>t)*...*P(Z_N>t)$$ Assuming i.i.d: $$P(T_1 \leq t) = 1-P(Z_1>t)^N = 1-(1-P(Z_1 \leq t))^N = 1-(1-F_{Z_1}(t)))^N$$ To find the PDF, we take the derivative of the CDF we just developed to get: $$\frac{d}{dt}P(T_1 \leq t)=N(1-F_{z_1}(t))^{N-1}*f_{z_1}(t)$$ Which is the exact formula you have.

Nevertheless, these assumptions aren't required but they do make our life easier.

Note: without loss of generality I used $Z_1$, however, you can choose any $Z_i$ since i.i.d.

Answer 2

Lack of independence makes this too complicated.

Independent but not identically distributed is still possible to analyse:

• The CDF of the minimum is $$F_{Z_{(1)}}(x) = 1- \prod\limits_i (1-F_{Z_i}(z))$$
• Taking the derivative and tidying up, the density of the minimum is $$f_{Z_{(1)}}(x) = \sum\limits_j\dfrac{f_{Z_j}(x)}{1-F_{Z_j}(z)} \prod\limits_i (1-F_{Z_i}(z))$$

As you would hope, this density becomes (1) from the question when the $Z_i$ are identically distributed.