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What does $0\le x,y\le1$ mean? See the red circled part in the image below for an example.

  1. I first thought it meant: $x\ge0$ and $y\le1$

  2. Then I thought it meant: $0\le x\le y \le 1 $

  3. But, based on the green part, I believe it means: $x$ and $y$ are in $[0,1]$

Is this notation unambiguous? In probability, the comma means $\bigcap$, so, to me, these are "separate" statements, as in (1) and not (3).

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    $\begingroup$ I would also interpret is as (c). $\endgroup$ – Malcolm Dec 10 '17 at 0:03
  • $\begingroup$ Thanks for your help guys. If you were to write out "a", how would you properly do it? Maybe $0 \le x \le \infty$, $-\infty \le y \le 1$ $\endgroup$ – HJ_beginner Dec 10 '17 at 0:07
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    $\begingroup$ $0\leq x \text{ and } y \leq 1$ $\endgroup$ – user326210 Dec 10 '17 at 0:08
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    $\begingroup$ @HJ_beginner What you have written after your edit is also correct. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '17 at 0:12
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    $\begingroup$ Saying that the comma means and may not help. You could still interpret it as either (0 <= x) and (y <= 1) or 0 <= (x and y) <= 1 :-) $\endgroup$ – user1324 Dec 10 '17 at 8:08
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I would take it as c, that both $x$ and $y$ are between $0$ and $1$ and think that it should be unambiguous. I might have a worry in my stomach that it was $a$ and be alert to the possibility as I read on or check back to make sure. I would say b is wrong and should be written the way you did.

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  • $\begingroup$ Thanks for your help! $\endgroup$ – HJ_beginner Dec 10 '17 at 0:28
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    $\begingroup$ It is a somewhat dangerous notation. If the spacing were different: $$0\le x,\quad y\le 1$$ you could think it was a. However, a would more often be written as $$x\ge 0,\quad y\le 1$$ $\endgroup$ – Jeppe Stig Nielsen Dec 10 '17 at 18:03
  • $\begingroup$ I would only ever have read that as $x\geq 0 $ *and* $y\leq 1$. $\endgroup$ – Therkel Dec 11 '17 at 14:19
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It is a convention for $0\leq x \leq 1$ and $0 \leq y \leq 1$ and is mostly to avoid typing it out twice. The comma is used this way in the equivalent statement $x,y \in [0,1]$ as well so it's consistent with that notation.

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The overall integral of $f_{X,Y}(x,y)$ has to be one. Let's then compare the different assumptions.

(a)

The integral $$\int_0^{\infty}\int_{-\infty}^1x+\frac32y^2\ dy\ dx$$

is not convergent.

(b)

$$\int_0^{1}\int_{x}^1x+\frac32y^2\ dy\ dx=\int_0^1x+\frac12-x^2-\frac12x^3\ dx=\frac{13}{24}\not=1.$$

(c)

$$\color{green}{\int_0^{1}\int_{0}^1x+\frac32y^2\ dy\ dx=\int_0^1x+\frac12\ dx=1}.$$

So, interpretation (c) seems to be correct.

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  • $\begingroup$ I appreciate you pointing that out. You're right I was getting very strange answers because I misinterpreted the region of $x$ and $y$. I was getting frustrated until I realized I was reading the problem wrong. Thanks. $\endgroup$ – HJ_beginner Dec 10 '17 at 0:45

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