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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.

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  • $\begingroup$ If there is independence then it is not difficult to find an expression for the CDF of $\min(Z_i)$. Taking its derivative gives the PDF. That gives formula $(1)$ if the $Z_i$ are moreover identically distribution. Follow this route and see for yourself how non-independence and non-identically distributed form obstacles. $\endgroup$
    – drhab
    Jul 1 at 8:38
  • $\begingroup$ look here for a counter exemple $\endgroup$ Jul 1 at 12:16

2 Answers 2

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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.

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    $\begingroup$ showing that iid is used for the formula does'nt proves that without idd assumption the formula does not hold .. (in fact it certainly does not, but in order to prove it, it could be more correct to show a counter exemple) $\endgroup$ Jul 1 at 8:59
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    $\begingroup$ @MrSmithGoesToWashington You are right in stating that this is not a proof. Nevertheless IMV it makes clear enough that "these assumptions are critical for this formula". In order to make things clear a proof is not always needed. $\endgroup$
    – drhab
    Jul 1 at 9:22
  • $\begingroup$ You are right my wording is misleading, I'll edit my answer. $\endgroup$
    – Aa me
    Jul 1 at 9:37
  • $\begingroup$ @ Aa me Thank you so much for your answer. $\endgroup$
    – chaaru
    Jul 1 at 14:32
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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.

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