1
$\begingroup$

I have this problem which I would like to discuss whether my solution is accurate. There are no solved examples like this one in my text book, so I want to ask anyone if they think my solution is correct or not.

"Let $X$~ $unif(0,1)$. Find the pdfs of the following random variables (be careful with ranges): (b) $Y=1/\sqrt{X}$."

I begin to note that because $X$ ~ $unif(0,1)$ we have that the cdf is $F_X(x) = x$, for $x \in [0,1] $.

Then we can express the cdf for $Y$ via

$F_Y(x) = P(Y \leqq x) = P(\frac{1}{\sqrt{X}} \leqq x) = ...= 1-P(X \leqq \frac{1}{x^2}) = 1-F_X(\frac{1}{x^2}) = 1-\frac{1}{x^2}$

Now it should be pretty straight forward, but because of the hint in the problem formulation I get a bit unsure in my reasoning:

We can easily differentiate the expression above to get the pdf for $Y$. In doing so, we note that we obtain a pdf $f_Y(x) = 2x^{-3}$. By definition, the pdf $f(x) \geqq 0 \forall x$ which is clearly not the case right now. So a restriction of the function to let it be defined such that

$f_Y(x) = 2x^{-3} , x > 0$ and zero elsewhere, would meet this requirement

However, this doesnt meet the requirement that $\int_{- \infty}^{\infty} f_Y(x) dx = 1$.

If we do, however, choose to restrict the function even further such that

$f_Y(x) = 2x^{-3} , x > 1$ and zero elsewhere,

this condition is fulfilled and thus $f_Y(x)$ is a pdf to $Y$. And this is the right answer according to my text book. But i wonder, is there something faulty with my arguments above? Is there a better way of solving this?

Thanks!

$\endgroup$
1
$\begingroup$

Your approach is fine but it can be shortened very much.

The restrictions are natural: if $X$ takes values in $(0,1)$, $Y=\frac{1}{\sqrt{X}}$ takes values in $(1,+\infty)$.
By denoting as $f_X$ and $f_Y$ the PDFs of $X$ and $Y$ we have $f_X(u) = \mathbb{1}_{(0,1)}(u)$ and

$$\forall a>1,\qquad \int_{1}^{a}f_Y(u)\,du = \mathbb{P}\left[\frac{1}{\sqrt{X}}\leq a\right]=\mathbb{P}\left[X\geq \frac{1}{a^2}\right]=1-\frac{1}{a^2} $$ so $f_Y(u) = \frac{2}{u^3}\mathbb{1}_{(1,+\infty)}(u)$ by differentiation.

$\endgroup$
  • $\begingroup$ Of course! I dont know why I didnt saw this right away. Thanks! $\endgroup$ – tarkovsky123 Apr 4 '18 at 17:01
1
$\begingroup$

Hint: if $X$ takes values between $0$ and $1$, where does $Y$ take values? You're correct that the pdf of $Y$ is $2 y^{-3}$ for some set of $y$, but it's not for all $y > 0$. Another way to think about this: for what $x$ is your statement $$1 - F_X(1/x^2) = 1 - 1/x^2$$ valid?

$\endgroup$

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.