Find cdf from pdf f(x) = [(k+1)x^k]/a^(k+1) 0

Question

Find cdf from pdf f(x) = [(k+1)x^k]/a^(k+1)  0<=x<=a
please help to check, I did some calculation in reply!

anonymous · Answer

$$f(x) = \frac{ (k+1)x^{k} }{a^{k+1} }  \rightarrow  0 \le x \le a    $$
so 
$$F(x)=\int\limits_{0}^{a} \frac{ (k+1)x^{k} }{ a^{k+1} } dx = \frac{ k+1 }{ a^{k+1} } \int\limits_{0}^{a}x^{k}dx$$
then
$$\frac{ k+1 }{ a^{k+1} } \times \frac{ x^{k=1} }{k+1 } ]_{0}^{a} \rightarrow \frac{ k+1 }{ a^{k+1} } \times (\frac{ a^{k+1} }{ k+1 } - \frac{ 1 }{ k+1 }) $$
then
$$F(x) = 1- \frac{ 1 }{a^{k+1} }$$

ybarrap · Answer

You are on the right track, but what you've calculated is the CDF for the entire domain, which should equal to 1. The last term in line 3 above should be zero, and then you can see that you will have 1 left.

Instead of "a" as the upper limit, use some random variate, call it y so that P( $\bf X$ $\le y$)=F(y). 

You are almost there, just make that change in line 3 and replace a with y.

anonymous · Answer

Thanks. I got it.
$$F(x) = (\frac{ x }{ a })^{k+1} $$