0
$\begingroup$

I'm recently started learning basic calculus and statistics, but I'm having a really hard time with some of the problems that I've been given (I come from a social science with no math background). In an problem that I've been trying to solve, I'm supposed to find the CDF of the following PDF:

$f_X(x) = \alpha \left( \frac{e^x}{(1+e^x)^2}\right)\left( \frac{e^x}{(1+e^x)}\right)^{\alpha-1}, \alpha>0$

I understand that I have to integrate this function in order to get the CDF, but I've only managed to do this:

$F_X(x) = \int \alpha \left( \frac{e^x}{(1+e^x)^2}\right)\left( \frac{e^x}{(1+e^x)}\right)^{\alpha-1}dx=\int \alpha\left( \frac{1}{(1+e^x)}\right) \left( \frac{e^x}{(1+e^x)}\right)\frac{\left(\frac{e^x}{(1+e^x)}\right)^{\alpha}}{\left(\frac{e^x}{(1+e^x)}\right)}dx=\int \alpha \left( \frac{1}{(1+e^x)}\right)\left( \frac{e^x}{(1+e^x)}\right)^\alpha=\int \left(\frac{e^x}{(1+e^x)^2}\right)^{\alpha+1} $

I'm not really sure if it's ok so far, and if it is I really don't know where to go from here. Please, if ny of you could give me some guidance it would be really helpful!

$\endgroup$

2 Answers 2

0
$\begingroup$

Hint: Notice that $$\int \alpha \left(\frac{e^x}{(1+e^x)^2}\right) \left(\frac{e^x}{1+e^x}\right)^{\alpha-1} dx=\int \alpha \left(\frac{e^x}{1+e^x}\right)^{\alpha-1} d\left(\frac{e^x}{1+e^x}\right)$$

So you can continue with the integration easier by focusing on $\int \alpha \cdot y^{\alpha-1}\ \ dy$ first.

$\endgroup$
0
$\begingroup$

The expression has been somewhat laid out in a way to help you.

Consider

$$\left(\frac{e^x}{e^x+1}\right)'=\frac{e^x(e^x+1)-e^xe^x}{(e^x+1)^2}=\frac{e^x}{(e^x+1)^2}.$$

Then

$$\left(\left(\frac{e^x}{e^x+1}\right)^\alpha\right)'=\alpha\frac{e^x}{(e^x+1)^2}\left(\frac{e^x}{e^x+1}\right)^{\alpha-1}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .