Positive continuous random variable. Determining he c.d.f and p.d.f

Question

Let X be a positive continuous random variable with density f$_x$(x)=1/x$^4$.

Answer 1

Hint:

You can find the PDF for Y using transformation.

https://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables

When you have the PDF for Y you can integrate to find the CDF.

Answer 2

My comments about the PDF just have to do with the fact that the problem is ill-posed as written in the sense that $f_X(x)=x^{-4}$ is not a valid PDF on $(0,\infty)$ or $(1,\infty)$, for instance, and so the problem cannot continue from here.

Change of variables: Let's work with $f_X(x)=3x^{-4}$ on $[1,\infty)$. Then this is a valid PDF ($f\geq 0$ and $\int f=1$). The CDF is then $\int_1^x 3u^{-4}du=1-x^{-3}$ for $x>1$.

I will sketch out the change of variables technique for you.

Let $Y=\log(X)$. To find $f_Y(y)$, let $h$ be any bounded measurable function on $\mathbb{R}$, then we first consider the integral, $$\mathbb{E}(h(Y))=\int h(y)f_Y(y) dy=\int_1^\infty h(\log(x)) f_X(x) dx=\mathbb{E}(h(\log(X)),$$ using the law of unconscious statisticians. Now use the subtitution $y=\log x$, which implies $x=e^y$, and since $1<x<\infty$, we have $0<y<\infty$ and finally $e^y dy =dx$. Proceeding, with the substitution, $$\int_1^\infty h(\log(x)) f_X(x) dx=\int_0^\infty h(y) f_X(e^y) e^y dy$$ $$=\int_0^\infty h(y) 3e^{-3y}dy$$ but since this holds for all bounded measurable functions $h$ we must have by identification that $f_Y(y)=3e^{-3y}$. One can check that this is a valid PDF on $(0,\infty)$ and further recognize it as the PDF of an exponential random variable with rate parameter $\lambda=3$ (and so mean $1/3$, etc).

Comment for further clarifications or questions that would not warrant a new post. If this change of variable technique is unfamiliar, peruse the link CruZ provided or find a calculus book and look up "change of variable" or "substitution" or "change of variable technique" or "transformations of random variables" in a probability textbook.

EDIT Here we derive the CDF of $Y$ first then differentiate to obtain the PDF of $Y$. We still use $f_X(x)=3x^{-4}$ on $(1,\infty)$. First, since, $1<X<\infty$ and $Y=\log X$, we have $$Y\leq y \iff \log X \leq y \iff X\leq e^{y}.$$ So that upon taking probabilities, $$\mathbb{P}(Y\leq y)=\mathbb{P}(X\leq e^{y}),$$ or in terms of CDFs, $$F_Y(y)=F_X(e^{y})$$ and now all you have to do is plug in $x=e^{y}$ into $F_X(x)=1-x^{-3}$ and this gives you $F_Y(y)$. Finally, differentiate with respect to $y$ to get $f_Y(y)=F_Y'(y)=3e^{-3y}$. These last steps I have purposely left out detail because it's algebra and calculus at this point and not probability proper. Go through the steps yourself to verify it works and practice what will eventually become routine computations. Beware that the CDF technique requires that the function $Y=g(X)$ is invertible and thus cannot be applied blindly (here $y=g(x)=\log x$ and has inverse $g^{-1}(y)=e^y$).