3
$\begingroup$

If $X \sim \text{U}(0,1)$, what is the distribution of $Y = \max(X,1/X)$? I know for this particular problem, $Y = \max(X,1/X) = 1/X$, whose distribution can be easily attained directly. However, I've been trying another approach by starting with this

$$P(Y \leq y) = P(X \leq y,\frac{1}{X} \leq y)=P(\frac{1}{X}\leq y| X \leq y)P(X \leq y),$$

and I've been stuck. I would be appreciate if someone could show me how it could be proceeded. Thank you very much.

$\endgroup$
4
  • 5
    $\begingroup$ Is $X$ really Uniform(0,1)? Because in that case, max($X, 1/X$) is always $1/X$. $\endgroup$
    – Cliff AB
    Jun 25, 2018 at 3:43
  • $\begingroup$ Yes, I've mentioned that for this particular case, max(X,1/X) would be 1/X, but I'm trying a more general way to deal with this kind of problem. $\endgroup$
    – diidoobiib
    Jun 25, 2018 at 3:44
  • 1
    $\begingroup$ If you tried $X \sim U(0,2)$, you could come up with a nice solution that might generalize (depending on what you want to generalize too). $\endgroup$
    – Cliff AB
    Jun 25, 2018 at 3:53
  • $\begingroup$ @CliffAB so supposing that X~U(0,2), how can I derive the distribution of max(X,1/X)? $\endgroup$
    – diidoobiib
    Jun 25, 2018 at 4:04

1 Answer 1

4
$\begingroup$

As you have acknowledged in your question, this problem is trivial in the case where $0 < X < 1$. In the more general case (taking $\tfrac{1}{0} = +\infty$ for definiteness) you have:

$$Y = \max(X, \tfrac{1}{X}) = \begin{cases} \tfrac{1}{X} & & \text{if }X < -1 \\[6pt] X & & \text{if }-1 \leqslant X < 0 \\[6pt] +\infty & & \text{if } X = 0 \\[6pt] \tfrac{1}{X} & & \text{if }0 < X < 1 \\[6pt] X & & \text{if }X \geqslant 1 \\[6pt] \end{cases}$$

If you restrict this to the case where $X>0$ (which includes the last two categories), you have:

$$\begin{equation} \begin{aligned} F_y(y) \equiv \mathbb{P}(Y \leqslant y) &= \mathbb{P}(\max (X, \tfrac{1}{X}) \leqslant y) \\[6pt] &= \mathbb{P}(X \geqslant \tfrac{1}{y}, 0 < X < 1) + \mathbb{P}(X \leqslant y, X \geqslant 1) \\[6pt] &= \mathbb{P}(\tfrac{1}{y} \leqslant X < 1) + \mathbb{P}(1 \leqslant X \leqslant y). \\[6pt] \end{aligned} \end{equation}$$

And of course, in the case where $0<y<1$ this probability is zero, as expected.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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