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After a long time without having touch anything related to maths or statistics, I decided to give myself another chance. I am currently refreshing some concepts of density and distribution functions, and I'm now facing a little misconception. Briefly, I have this density function:

\begin{cases} 0 & \text x < 0\\ 2x/(1+x^2 )^2 & \text{x ≥ 0} \end{cases}

from which I have obtained the following distribution function after calculating the integrate:

\begin{cases} 0 & \text x < 0\\ -1/(1+x)^2 & \text{x ≥ 0} \end{cases}

My problem is that I'm trying to plot the distribution function using R, but since the distribution function is negative, I'm not sure if it is correct (shouldn´t it be positive?) and which would be the best way to plot it (of course, I'm a R newby too).

Thank you in advance.

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    $\begingroup$ You quote the indefinite integral, which is defined only up to an additive constant. For the definite integral you are trying to obtain, choose that constant appropriately. $\endgroup$
    – whuber ♦
    Commented Mar 31, 2019 at 16:21

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You've a typo in the denominator of your distribution function: $x$ should be $x^2$, w/o the square of the parentheses. And, as @whuber said, you need a constant term after the integration (for $x\geq 0$):

$$F(x)=\int \frac{2xdx}{(1+x^2)^2}=-\frac{1}{1+x^2}+C$$

By definition, $\lim_{x\rightarrow \infty}F(x)=1$, substituting yields $$\lim_{x\rightarrow\infty}F(x)=C-\lim_{x\rightarrow\infty}\frac{1}{1+x^2}=C=1$$

So, you can plot the following, for $x\geq 0$, (for $x<0$, $F(x)=0$): $$F(x)=1-\frac{1}{1+x^2}=\frac{x^2}{1+x^2}$$

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  • $\begingroup$ Thank you gunes, this helps a lot. But I have a (naive, maybe) question. The distribution function, isn´t it supposed to be the cumulative of the density function? I mean, for the density function, f(1)+f(2)=0.66, but for the distribution, according to your F(x), then F(2)=0.8. Shouldn´t these values be the same? $\endgroup$
    – Voronin_10
    Commented Mar 31, 2019 at 18:14
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    $\begingroup$ Yes, $F(x)$ is CDF and is actually defined as $P(X\leq x)=\int_{-\infty}^x f(x)$. You cannot sum individual $f$'s because $x$ is continuous. What about $f(1.5),f(1.55),f(1.553),...$, $\endgroup$
    – gunes
    Commented Mar 31, 2019 at 18:37

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