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I found this fun little problem intriguing as it really questions your assumptions.

Given two unknown probabilities, $p_1$ and $p_2$, along with the condition $p_1+p_2 \le 1$, what is the best estimate for $p_1$ and $p_2$? (Note the inequality)

Couple possible solutions, and due to symmetry all probabilities are equal

  • A) $p=\frac{1}{2}$. The inequality can be an equality and divided two ways
  • B) $p=\frac{1}{3}$. The inequality suggests there is a possibility of a third probability, and the most uncertain out of three probabilities is each equal 1/3.
  • C) $p=\frac{1}{4}$. Any value for the sum inside the inequality may be likely, so the expected value is $\frac{1}{2}$ so two probabilities summing to $\frac{1}{2}$.

I was leaning towards (B), but then realized that assumes the probabilities should add to one. The value of one in the inequality is numerically equal to the normalized sum of probabilities, but there's no reason the inequality should be one. What if we had $p_1+p_2 \le 1.5$ ? (A) now suggest $p=\frac{3}{4}$ while (B) suggests $p=\frac{1}{2}$.

For the general problem of $p_1+p_2 \le c$ with $c$ large enough we get into the upper limit of definition of probability $0 \le p_i \le 1$. With that reasoning, the answer to $p_1+p_2 \le 100$ should be the same as $p_1+p_2 \le 5$. Is that now leaning towards (A) with $p=\frac{1}{2}$ for large enough enough $c$, or it now turns into $p=1$?

Thought of a 'compromise' solution for large enough $c$, for $p_1 + \dots + p_n \le c$

$p=\frac{c}{n+1}$ for $0 \le c \le \frac{1+n}{2}$ and $p=\frac{1}{2}$ otherwise

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    $\begingroup$ This doesn't seem to ask a specific question but reads more like a chatty blog-post attempting to discuss a problem you find interesting (which would generally be unsuitable for a Q&A site). Can you be clearer about what your question is, and clearly distinguish that from the parts where you're describing your thinking? $\endgroup$ – Glen_b Apr 6 '16 at 5:26
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I think $\frac{1}{3}$ is the best answer.

All we know is that $x+y\leq1$, so I think we should assume a uniform distribution within those bounds

Since the pdf has an area of $\frac{1}{2}$ with non-zero values $f_{xy}(x,y)=2$, $x+y\leq1$, $x,y\geq0$

$E[x]=\int_{xy}{x f_{xy}(x,y)}=\int_{x=0}^{1}{\int_{y=0}^{1-x}{2x}dx}dy=1-\frac{2}{3}=\frac{1}{3}$

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    $\begingroup$ I can see no basis in the question to justify any distributional assumption on the $p_i$. $\endgroup$ – whuber Apr 6 '16 at 13:51

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