1
$\begingroup$

Let $U$, $U_1$ and $U_2$ be independent uniform random numbers between 0 and 1. Can we show that generating random number $X$ by $X = \sqrt{U}$ and $X = \max(U_1,U_2)$ are equivalent?

$\endgroup$
1
$\begingroup$

Random variables $X = \sqrt{U}$ and $Y = \max(U_1, U_2)$ are equal in distribution. Indeed, both $0 \leqslant X(\omega) \leqslant 1$ and $0 \leqslant Y(\omega) \leqslant 1$. Furthermore, for $0 \leqslant x \leqslant 1$, we have $$ F_X(x) = \mathbb{P}(X \leqslant x) = \mathbb{P}(\sqrt{U} \leqslant x) = \mathbb{P}(U \leqslant x^2) = x^2 $$ $$ F_Y(x) = \mathbb{P}(Y \leqslant x) = \mathbb{P}(\max(U_1,U_2) \leqslant x) = \mathbb{P}(U_1 \leqslant x, U_2 \leqslant x) \stackrel{\text{independence}}{=} \\\mathbb{P}(U_1 \leqslant x) \cdot \mathbb{P}(U_2 \leqslant x) = x^2 $$

$\endgroup$
1
$\begingroup$

For every $x$ in $(0,1)$, $\mathrm P(\max\{U_1,U_2\}\leqslant x)=\mathrm P(U_1\leqslant x)\cdot\mathrm P(U_2\leqslant x)=x\cdot x=x^2$ and $\mathrm P(\sqrt{U}\leqslant x)=\mathrm P(U\leqslant x^2)=x^2$ hence $\max\{U_1,U_2\}$ and $\sqrt{U}$ follow the same distribution.

$\endgroup$

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.