Derivation of P.D.F. from distribution function

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SUMMARY

The discussion centers on the derivation of the probability density function (PDF) from the distribution function for a random variable. The author successfully computed the PDF, denoted as ##f_{X_k}(x)##, using the cumulative distribution function (CDF) ##F_{X_k}(x)##, expressed as ##F_{X_k}(x) = p(X \leq x) = \displaystyle\sum_{j=k}^n \binom{n}{j} F^j(x)(1-F(x))^{n-j}##. The final result is confirmed as ##f_{X_k}(x) = \frac{n!}{(k-1)!(n-k)!}f(x) F^{k-1}(x)(1-F(x))^{n-k}##, demonstrating a clear understanding of the computations involved.

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WMDhamnekar
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TL;DR
If ## F_{X_k}(x) = p(X \leq x) = \displaystyle\sum_{j=k}^n \binom{n}{j} F^j(x)(1-F(x))^{n-j}, -\infty
< x < \infty ## then how to prove ##f_{X_k} (x) =\frac{n!}{(k-1)!(n-k)!}f(x) F^{k-1}(x)(1-F(x))^{n-k}##
Author computed ##f_{X_k}(x)## as follows but I don't understand it. Would any member explain me the following computations?
1655626767982.png
 
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WMDhamnekar said:
Summary: If ## F_{X_k}(x) = p(X \leq x) = \displaystyle\sum_{j=k}^n \binom{n}{j} F^j(x)(1-F(x))^{n-j}, -\infty
< x < \infty ## then how to prove ##f_{X_k} (x) =\frac{n!}{(k-1)!(n-k)!}f(x) F^{k-1}(x)(1-F(x))^{n-k}##

Author computed ##f_{X_k}(x)## as follows but I don't understand it. Would any member explain me the following computations?
1655659126417.png
I tag this question as "SOLVED". I understood all the computations. Thanks.
 
Last edited:

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