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Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$

so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.

Probability that $x<2$ would be $\frac{1}{2}$ as $x$ can be in $(1,2)$ where sample set of $x$ is $(1,3)$, while probability that $x^2<4$ is $\frac{3}{8}$ as $x^2$ can be in $(1,4)$ where sample set of $x^2$ is $(1,9).$

So why is the probability that $x<2$ different from $x^2<4$ if both are identical?

Juniven Acapulco
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kuuhaku
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    The random variable $X^2$ is not also uniformly distributed. – Leon Sot Jan 02 '17 at 12:29
  • As Leon pointed out, $X^2$ is not uniform. Detail here. – msm Jan 02 '17 at 12:37
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    But it is nevertheless a very nice question, as nobody has pointed out as of yet. – barak manos Jan 02 '17 at 12:40
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    @barakmanos: it's gotten 4 upvotes so far. I'd say that at least 4 people have pointed that out ;-) – robjohn Jan 02 '17 at 12:46
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    @robjohn: Yep, but the first one was mine :) ... BTW, although the answer is simple, I'm having a hard time finding a simple way to explain it. – barak manos Jan 02 '17 at 12:47
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    Also, for whoever chooses to down-vote this question, please leave a comment explaining why, as there doesn't seem to be any reason to do so. The question is nicely phrased, and shows effort in digging into the problem. – barak manos Jan 02 '17 at 12:56
  • @msm - why is the most upvoted answer here different from the link you pointed out? – NindzAI Oct 17 '17 at 02:37
  • it is the same, the link explains Leon's statement while the answer explains my doubt. – kuuhaku Oct 19 '17 at 09:44

2 Answers2

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If $X$ is uniformly distributed on $[1,3]$, then $P(X^2\le a^2)=P(X\le a)=\frac{a-1}2$. Therefore, $P(X^2\le a)=\frac{\sqrt{a}-1}2$. Thus, the PDF of $X^2$ is $\frac1{4\sqrt{a}}$.

That is, $X^2$ is not uniformly distributed on $[1,9]$.

robjohn
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You decided for any $x$ to match $x^2$, so you are claiming that $x<2$ is identical to $x^2<4$, but you can also match $4x-3$ in the interval $(1,9)$ for all $x$ in the interval $(1,3)$, which makes $x<2$ the same as $4x-3<5$.

That is the problem of defining probability on infinite group.

76david76
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  • x<2 is identical to x^2<4 in any way you look at it. btw i got my mistake that x^2 is not uniformly distributed. – kuuhaku Jan 02 '17 at 13:09