Find $f_W$ for $W=-\frac{1}{2}\log(X)$. Given answer doesn't match my answer

Question

Let $X$ a random variable with the p.d.f.

$$ f_X(x) = \begin{cases} 3x^2, & \text{if } 0<x<1 \\ 0, & \text{otherwise} \end{cases}$$

and $W=-\frac{1}{2}\log(X) $. Find the p.d.f. $f_W$ for $W$.

I've been given this task and the answer to the question as well. So, my idea was to use the formula:

$$f_Y(y)=\frac{f_X(x_1)}{g'(x_1)}$$

where $\ g(x_1)=y$.

However, the answer that I am ending up with simply does not align with the answer given by the instructor. I am told that the answer should be $W \sim \operatorname{Exp}(6) $.

This is how I'm breaking the task down:

$$ f_X(x) = \begin{cases} 3x^2, & \text{if } 0<x<1 \\ 0, & \text{otherwise} \end{cases} \qquad \text{and} \qquad g(x) = -\frac{1}{2}\log(x)$$

are given by the task. So what I need to do is differentiate $g(x)$ and find the inverse function of $W$ and put it all together:

• Derivative: $g'(x) = -\frac{1}{2x} $.
• Inverse: $x_1 = g^{-1}(x) = e^{-2x}$.

Putting it together:

$$ f_W(x) = \frac{3 \cdot (e^{-2x})^2}{-1/(2 \cdot e^{-2x})} =-6(e^{-2x})^3 $$

As you can see this doesn't match the given answer. I've used hours trying other methods but cannot find the mistake. Any thoughts?

Best regards

Answer 1

The cdf of $W$ can be calsulated as follows

$$F_W(w)=P(W<w)=P\left(-\frac12\log(X)<w\right)=$$ $$=P\left(\log(X)>-2w\right)=P\left(X>e^{-2w}\right)=$$ $$=3\int_{e^{-2w}}^1x^2\ dx=\begin{cases}0&\text{ if } &w<0\\1-e^{-6w}&\text{ if } &0\leq w.\end{cases}$$

The density is then

$$f_W(w)=\frac{d F_W}{dw}=6e^{-6w}$$

if $w\geq 0$.

Answer 2

The pdf of $W$ can be written as $$ f_W(w)=\int_0^1 dx\ 3x^2\delta\left(w+\frac{1}{2}\ln x\right)\ , $$ using a Dirac delta. Since $w+(1/2)\ln x=0\Rightarrow x=e^{-2w}$, we have $$ f_W(w)=\int_0^1 dx\ 3x^2\frac{\delta\left(x-e^{-2w}\right)}{\frac{1}{2e^{-2w}}}=6e^{-2w}\times e^{-4w}\mathbb{1}(e^{-2w}<1)=6e^{-6 w}\theta(w)\ , $$ which is correctly normalized over $w\in (0,\infty)$. The fact that the support is $(0,\infty)$ is evident from the definition, as the $\log$ of a number $X$ between $0$ and $1$ is negative. I think all you are missing is the absolute value of $g'(x)$.

PS I highly doubt that your instructor gave you $w\sim e(6)$ as a solution. That notation doesn't make any sense, to start with...

Answer 3

Let $w\in (0,\infty)$. Let $F_W$ be the CDF of of $W$ and $F_X$ the CDF of $X$

\begin{align} F_W(w)&=P(W<w)=P\left(\frac{-\ln(X)}{2}<w\right)\\ &=P(\ln(X)>-2w)=P(X>e^{-2w})\\& =1-P(X\leq e^{-2w})=1-F_X(e^{-2w}) \end{align} Differentiating and using the Fundamental Theorem of Calculus gives us the pdf of $W$. So: \begin{align} f_W(w)=2e^{-2w}f_X(e^{-2w})=2e^{-2w}3e^{-4w}=6e^{-6w} \end{align} since $w\in(0,\infty)$. It can be easily seen that $f_W(w)=0$ for $w\leq 0$.