0
$\begingroup$

The random variable X has probability density function

\begin{cases} \frac{x}{8} & 0<x<4 \\ 0 & otherwise \end{cases}

Find the probability density function of $Z = \log_e (X/4)$

I tried and got this:

\begin{cases} \frac{4e^z}{8} & 0<Z<e^z \\ 0 & otherwise \end{cases} but the limits don't seem right to me, is this correct?

$\endgroup$
  • 1
    $\begingroup$ Hint: what values can $Z$ get for $0<x<1$? $\endgroup$ – karakfa Nov 16 '16 at 20:32
  • $\begingroup$ would it be $0<Z<\infty$? $\endgroup$ – charliejs Nov 16 '16 at 20:45
  • $\begingroup$ @charliejs ?? How do you realize $Z=42$ from some $X$ in $(0,4)$ through the transform $Z=\ln(X/4)$? $\endgroup$ – Did Nov 16 '16 at 21:03
  • $\begingroup$ @Did I haven't got $Z=42$?? $\endgroup$ – charliejs Nov 16 '16 at 21:10
  • $\begingroup$ You suggested that the range of $Z$ is $(0,\infty)$, right? And $42$ is in $(0,\infty)$, yes? So... $\endgroup$ – Did Nov 16 '16 at 21:16
2
$\begingroup$

if $f(x)$ is the profability density function for $x$ and $z = z(x)$, then the pdf for $z$ can be calculated as

$$ f(z) = f(x)\left|\frac{dx}{dz}\right| \tag{1} $$

In this case you already know the expression $z=z(x)$:

$$ z = \ln x/4 \quad\Rightarrow\quad x = 4e^{z} $$

so that

$$ \frac{dx}{dz} = 4e^z $$

Before replacing in (1) note that when $z(x\to 0^+) = -\infty$, and $z(x=1) = 0$, so the $z$ will range in $(\infty, 0)$. So in the this range

$$ f(z) = f(x)|4e^z| = 2e^{2z} \quad\mbox{for}\quad z<0 $$

or in other words

$$ f(z) = \left\{\begin{array}{lcl} 2e^{2z} &,& z < 0 \\ 0 &,& {\rm otherwise}\end{array}\right. $$

Here's a simulation, the dashed line is the function $2e^{2z}$

enter image description here

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.