Calculating CDF of Max of IID Random Variables with CDF F(x) and PDF f(x)

[ekaveera100]

1. X_1,X_2\cdots X_n\:\text{are IID Random Variables with CDF}\,F(x)\:\text{and PDF}\,f(x)\\<br /> \text{then What is the CDF of Random variable }\,Max(X_1,X_2\cdots X_n)

Homework Equations

3. \text{Since Y will be one among}\,X_1,X_2\cdots X_n,\text{why cannot its CDF be }\,F(x)\\\text{I need to know flaw in my answer}

 

[LCKurtz]

1. X_1,X_2\cdots X_n\:\text{are IID Random Variables with CDF}\,F(x)\:\text{and PDF}\,f(x)\\<br /> \text{then What is the CDF of Random variable }\,Max(X_1,X_2\cdots X_n)

3. \text{Since Y will be one among}\,X_1,X_2\cdots X_n,\text{why cannot its CDF be }\,F(x)\\\text{I need to know flaw in my answer}

Intuitively, here's what's wrong with that. Take the simpler case of n IID random variables $X_1,\ X_2,...X_n$ uniformly distributed on [0,1]. If you take a samples $x_1,\ x_2,...x_n$ from these distributions, and you always choose the largest value, wouldn't you expect your answer to be biased towards the larger numbers in the interval? Suppose you take 20 samples and consider the largest value. It would be very unlikely for the max to be less than 1/2, wouldn't it? $(\frac 1 2)^{20}$ to be exact, even though each sample had an a priori probability 1/2 of being less than 1/2.