0
$\begingroup$

Exercise

An exercise asked me, given a density function $[h(x) = 1/x^2 \Longleftrightarrow x>1]$ of a continuous random variable X, will find and represent the distribution function $ H_X $.

Solution

Then I calculated $ H_X $ and gave me $ - \dfrac{1} {x} + C $, but the function is different others:

enter image description here

Did I do something wrong? Or what is happening?

1 Answers 1

1

If the density is $$\frac1{x^2}, \ x\ge 1$$

Then the distribution function at an $y$ is

$$F(y)=\int_1^y \frac1{x^2}\ dx=-\left[\frac1x\right]_1^y=1-\frac1y, \ y\ge 1.$$

What is wrong with this?

So, you have a definite integral and not a indefinite one!

The following figure depicts the graph of the density and that of the distribution function.

enter image description here

The blue line is the density and the purple line is the distribution. At an $y$ the value of the purple line tells the area below the blue line from $1$ to that $y$ -- in accordance with the definition of the distribution function and that of the density

$$P(X

  • 0
    Being accustomed to the fact that the lower bound was always zero, and that the primitive always gave zero, I omitted Barrow's rule. So the result did not give me. Thank you. – 2017-02-12
  • 1
    Thank you. I've added a last sentence to my answer. Pls. read it. – 2017-02-12