# Limits of Integration Cumulative Distribution Function

Not sure if this is just some integration "trickery" which I don't grasp, or if I'm missing something on the theory side of distributions.

Why is the upper limit of integration set to x?

$$f_{X}(x)=\left\{\begin{array}{ll}{\frac{2}{a}-\frac{2}{a^{2}} x,} & {0

$$F_{X}(x)=\int_{0}^{x}\left\{\frac{2}{a}-\frac{2}{a^{2}} t\right\} d t=\left[\frac{2}{a} t-\frac{1}{a^{2}} t^{2}\right]_{0}^{x}=\frac{2}{a} x-\frac{1}{a^{2}} x^{2}, \quad 0 \leq x \leq a$$

Why is this formulation not defined , with $$a=0 , b=a$$?

$$\mathrm{P}(a

The question mentions the fact that $$a$$ is a positive constant. I understand the significance of this with regards to $$a \neq 0$$, since cfd is not defined for any specific value of $$x$$ given the continuous nature of the random variable. But I don't understand how $$a$$ not being a defined integer changes the calculation. Isn't $$a$$ still just the upper bound such that for $$x > a , P(X < x) = 1$$

What exactly is the difference between the above, and the following?

$$\begin{array}{c}{f(x)=3 x^{2} \text { on }[0,1] \Rightarrow F(y)=\int_{0}^{y} 3 x^{2} d x=y^{3} \text { on }[0,1] . \text { Therefore, }} \\ {\qquad F(y)=\left\{\begin{array}{ll}{0} & {\text { if } y<0} \\ {y^{3}} & {\text { if } 0 \leq y \leq 1} \\ {1} & {\text { if } 1

• "Isn't a still just the upper bound such that for $x>a$,$P(X<x)=0$"... I'd say $P(X<x)=1$. – dfnu Dec 2 at 21:28
• Indeed, thanks for that! – Joao Almeida-Domingues Dec 2 at 21:31
• As for the rest, I see some confusion, in the role played by the letter $a$. In the first example $a$ is a parameter of the distribution. So, depending on $a>0$ the distribution (also its support) changes. In your second case, you are using $a$ as the independend variable (limit of intergration) to obtain CDF from PDF. – dfnu Dec 2 at 21:45
• I see the confusion, but my point was regarding the interval of validity/support. More generally, perhaps, why is the upper bound set to the independent variable on both? – Joao Almeida-Domingues Dec 2 at 21:53
• That is because CDF is defined as $P(X\leq x)$. So if you have PDF $f_X(x)$ you can obtain CDF as $$F_X(x) = \int_{-\infty}^x f_X(t) dt.$$ – dfnu Dec 2 at 21:54