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My question is with reference to wireless communication.

There are $N$ independent and identically distributed random channels i.e., $Z_i \in \{ Z_1, Z_2,\ldots, Z_N\}$ 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

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

Will equation (1) will be valid if the random channels are not independent and not identically distributed?

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  • $\begingroup$ I tried to improve your title. Your question is actually not very much related to ordered statistics, but simply to basic stochastics! $\endgroup$ Jul 1 at 9:28
  • $\begingroup$ (I also improved your formula's look and how the (1) is added, like I did for a previous question for you. It might be cooler if you reviewed the changes happening to your questions and applied them yourself in your next question :) A title called "Related to..." almost certainly is not as specific and clear as it could be.) $\endgroup$ Jul 1 at 9:38

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If all you want is a yes or no answer -- no.

The proof is easy: in the extreme case, let each $Z_i = Z$. Then the pdf of the minimum is just the pdf of $Z$, and that doesn't equal your (1).

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  • $\begingroup$ Thanks a lot sir for your answer to the point answer. $\endgroup$
    – chaaru
    Jul 3 at 3:09
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No, it will not be valid. Think about where the exponent comes from. Also, if you have different distributions, how can there be only an arbitrarily picked $f_{Z_i}$ in your (1)? That doesn't make sense. You should probably build up a little more intuition here!

If you're working on ordered statistics, calculating the PDF of $Z_\min=\min(\{Z_i\})$ for i.i.d. RVs is probably something you want to do at least once yourself to get yourself warm with the common tricks when dealing with collections of random variables.

The derivation of $f_{Z_\min}$ is actually pretty straightforward. I'll give you a start!

\begin{align} f_{Z_\min}(z) &=\frac{\mathrm d}{\mathrm d\,z}F_{Z_\min}(z) \tag2\\ &=\frac{\mathrm d}{\mathrm d\,z}P({Z_\min}\le z)\tag3\\% &=\frac{\mathrm d}{\mathrm d\,z}P(Z_1\le z \,\wedge\, Z_2\le z \,\wedge\,Z_3\le z \,\wedge\,\ldots\,\wedge\,Z_N\le z )\tag4\\% \end{align}

Now, at this point your calculation starts depending on whether $Z_i$ are independent! Try with independent variables $Z_i, \quad i=1,\ldots,N$ first. Remember the definition of what it means for random variables to be independent. Then, don't forget to apply the chain rule of derivatives.

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  • $\begingroup$ Thank you so much sir for your detailed answer. And I will also improve my question asking capability :) $\endgroup$
    – chaaru
    Jul 1 at 14:29

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