Solving cumulative distribution function from exponent distribution function by integration Exponent probability density function would be defined as:
$$ f(x)= \begin{cases} \lambda e^{-\lambda x}, & x \ge 0. \\
 0, & x\le 0.
 \end{cases}$$
Now if i want cumulative distribution function from this i integrate this from $- \infty$ to $\infty$? this should be:
$$F(x)=\int_{-\infty}^x f(s) \, ds=\begin{cases} 1-e^{-\lambda x}, & x> 0. \\
0, & x\le 0\end{cases}$$
Now there are few things i dont understand. Why we are integrating with different variable than we originally had ? and why we are using $x$ as upper limit ? shouldn't this be $\infty$.
Also if someone could explain all the steps between defining integration and end result. Since i've been trying integrate this by hand but it doesn't seem i would get correct result.
thanks,
Tuki
 A: The point of a probability density function (lower-case) $f$ of a random varible (capital) $X$ is that for any two numbers $a,b$ with $a<b$ we have
$$
\Pr( a\le X\le b) = \int_a^b f(s)\,ds.
$$
And the point of a cumulative probability distribution function (capital) $F$ of a random variable $X$ is that for any number $x$ we have
$$
F(x) = \Pr(X\le x).
$$
Putting these two ideas together we have
$$
F(x) = \Pr(X\le x) = \int_{-\infty}^x f(s)\,ds.
$$
The variable $s$ goes from $-\infty$ to $x$; it runs through the whole set of numbers that are less than or equal to $x.$ Thus it is not the same thing as $x.$ We could have called it $t,$ and then we have
$$
F(x) = \int_{-\infty}^x f(t)\,dt
$$
and that's just a valid, provided it's not in a context in which the letter $t$ refers to something else.
Logically it is like the following situation:
$$
\sum_{i=1}^3 i^2 = 1^2 + 2^2 + 3^2 = \sum_{j=1}^3 j^2.
$$
In the sum $1^2+2^2+3^2$ we don't see anything called $i$ or $j.$ In the first term, $1^2,$ we have $i=1$ or $j=1,$ and in the second term, $2^2,$ we have $i=2$ or $j=2.$ We can call that "bound variable" either $i$ or $j$ or $k$ or something else.
