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?

$$ f_{X}(x)=\left\{\begin{array}{ll}{\frac{2}{a}-\frac{2}{a^{2}} x,} & {0<x<a} \\ {0,} & {\text { otherwise }}\end{array}\right. $$

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

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

$$\mathrm{P}(a<X \leq b)=F_{X}(b)-F_{X}(a)=\int_{a}^{b} f_{X}(t) d t$$

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<y}\end{array}\right.}\end{array}

  • $\begingroup$ "Isn't a still just the upper bound such that for $x>a$,$P(X<x)=0$"... I'd say $P(X<x)=1$. $\endgroup$ – dfnu Dec 2 at 21:28
  • $\begingroup$ Indeed, thanks for that! $\endgroup$ – Joao Almeida-Domingues Dec 2 at 21:31
  • $\begingroup$ 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. $\endgroup$ – dfnu Dec 2 at 21:45
  • $\begingroup$ 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? $\endgroup$ – Joao Almeida-Domingues Dec 2 at 21:53
  • $\begingroup$ That is because CDF is defined as $P(X\leq x)$. So if you have PDF $f_X(x)$ you can obtain CDF as $$F_X(x) = \int_{-\infty}^x f_X(t) dt.$$ $\endgroup$ – dfnu Dec 2 at 21:54

Your Answer

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

Browse other questions tagged or ask your own question.