1
$\begingroup$

I have a function $f(x)=2ae^{-ax}(1-e^{-ax})$, for $x>0, a>0$. This is a pdf. I need to find $P(X>1)$. I have done all my work in such a way that I should get the same answer whether I use the pdf or the cdf to find this probability. However, I'm getting different answers. Can someone please help me?

My attempt:

(using pdf) $P(X>1)=\int_1^{\infty}2ae^{-ax}(1-e^{-ax})dx = 2e^{-a}-e^{-2a}$

(using cdf) $P(X>1)= 1-P(X\leq 1) = 1 - (F_X(1)) = 1-(e^{-ax}(e^{-ax}-2))|_{x=1}=1-2e^{-a}-e^{-2a}$

Why are my answers different? Thanks!

$\endgroup$
  • $\begingroup$ The first integral should start at 1 (maybe its just a typo) $\endgroup$ – Michael M Oct 11 '13 at 12:43
  • $\begingroup$ There's no such thing as a cumulative density function. The word "cumulative" contradicts the word "density". See this disambiguation page on Wikipedia: en.wikipedia.org/wiki/Cumulative_density_function $\endgroup$ – Michael Hardy Oct 11 '13 at 17:27
2
$\begingroup$

It looks like a simple calculation error:

\begin{align} F_X(y)&=\int_{x=0}^y 2ae^{-ax}(1-e^{-ax})\\ &=e^{-ax}(e^{-ax}-2)|_{x=0}^y\\ &=e^{-2ay}-2e^{-ay}+1. \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy