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My question is: Suppose $X_1,...X_n$ are independent random variables from a continuous function with common $CDF$ $ F_Y(y)$ and common $PDF $ $f_X(x)$. Let $ Y= max \lbrace X_1,...X_2 \rbrace $. Now I showed in part a) that $ F_Y(y) = (F_X(y))^n$, but I am stuck on part b). Part b) asks me to derive the PDF, $f_Y(y) $. Any suggestions????

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$$f_y(y) = \frac{dF_Y(y)}{dy} = \frac{dF_X^n(y)}{dy} = n F_X^{n-1}(y) f_X(y)$$

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  • $\begingroup$ But we had $ (F_X(x))^n $ on the right hand side which is in terms of x, not y... so do you just ignore the x and change it to y ? $\endgroup$ – MathStudent Nov 25 '13 at 23:02
  • $\begingroup$ @MathStudent You had it wrong. I fixed your question also. Otherwise the equation makes no sense - LHS in $y$ and RHS in $x$, but also the proof makes no sense either... $\endgroup$ – gt6989b Nov 25 '13 at 23:04
  • $\begingroup$ Are you sure? because this is what my professor posted :/ $\endgroup$ – MathStudent Nov 25 '13 at 23:06
  • $\begingroup$ @MathStudent The argument for $F_Y(y)$ in terms of $F_X$ is only possible if they evaluate the same thing :) otherwise how can you draw any conclusions? What you are really saying is $$ F_Y(y) = \mathbb{P}[Y \leq y] = \mathbb{P}[\max_{i=1}^n X_i \leq y] = \prod_{k=1}^n \mathbb{P}[X_i \leq y] = \prod_{k=1}^n F_X^n(y) = F_X^n(y). $$ $\endgroup$ – gt6989b Nov 25 '13 at 23:08
  • $\begingroup$ You have convinced me lol so I'll have to let my prof know. Thank you. $\endgroup$ – MathStudent Nov 25 '13 at 23:14

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