Finding cdf and pdf of variable.

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SUMMARY

The discussion focuses on finding the cumulative distribution function (CDF) and probability density function (PDF) of the minimum variable Y, derived from four independent random variables X1, X2, X3, and X4, each with the PDF f(x) = 3(1-x)² for 0 < x < 1. The initial approach incorrectly calculates P(Y ≤ y) using the product of probabilities, leading to confusion regarding the integration of the PDF. The correct method involves integrating the PDF to derive the CDF and subsequently the PDF of Y, which is confirmed to be (1-y)¹².

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Homework Statement



Let X1 X2 X3 and X4 be four independent random variables, each with pdf f(x) = 3(1-x)2, 0<x<1, zero elsewhere. If Y is the minimum of these four variables, find the cdf and pdf of Y.

The Attempt at a Solution



P(Y<or= y)

= 1 - P(Y>y)
= 1 - P(X1>y, X2>y, X3>y, X4>y)
= 1 - P(X1>y)P(X2>y)P(X3>y)P(X4>y)
= 1 - [3(1-x)2]4

which does NOT equal (1-y)12, which is the answer in the back of the book. Don't know where I'm going wrong...
 
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What is P(x_1 > y)? It's certainly not that. Maybe you should try integrating over f instead.
 
clamtrox said:
What is P(x_1 > y)? It's certainly not that. Maybe you should try integrating over f instead.

Got it! Thanks so much for the help.
 

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