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Let $U_1, U_2, U_3$ be independent random variables that are each distributed uniformly in $(0,1)$. What is the probability that the second highest value among them lies between $\frac{1}{3}$ and $\frac{2}{3}$.

Could someone help me solve this? Let X be the second highest of U1, U2, U3

I did some research and found the pdf of second order statistic amongst n uniformly distributing random vairables and it is

$f_{X(2)}(x) = \frac{n!}{(j-1)!(n-j)!}x^{j-1}(1-x)^{n-j} = 6x(1-x)$ when n = 3 and j = 2

Satish Ramanathan
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    You can find a derivation of the density function of $\max(U_1,U_2,U_3)$ here: https://stats.stackexchange.com/questions/18433/how-do-you-calculate-the-probability-density-function-of-the-maximum-of-a-sample – dcolazin Jul 18 '19 at 14:33
  • More generally, you can read about $\max$ here: https://en.wikipedia.org/wiki/Order_statistic – dcolazin Jul 18 '19 at 14:34
  • This one is a bit different. This does not asks you for max but the middle one of an increasing values that lie between the interval 1/3 and 2/3. Quite tricky!! – Satish Ramanathan Jul 18 '19 at 14:35
  • Yeah, the max is a special case of order statistic, but the ideas behind the computation are the same. – dcolazin Jul 18 '19 at 14:36
  • Thanks for giving me the clue to check for the distribution of order statistic. I did the research and got the answer. – Satish Ramanathan Jul 18 '19 at 15:19
  • The integral $\int_{1/3}^{2/3} 6x(1-x) : dx$ will give you the answer. Alternatively, you can solve this combinatorially - think about how $U_1, U_2, U_3$ are distributed into three bins $(0, 1/3), (1/3, 2/3), (2/3, 1)$. – Michael Lugo Jul 18 '19 at 15:24
  • and then.... how do you get to the answer. – Satish Ramanathan Jul 18 '19 at 15:26
  • @SatishRamanathan Since you found an answer, can you post it here (as an answer)? – Mars Plastic Jul 18 '19 at 15:55

2 Answers2

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Let events $A,B,C$ be defined as follows

  • $A$ is the event that at least two of $U_1,U_2,U_3$ are in the interval $[0,{\large{\frac{2}{3}}})$.$\\[6pt]$
  • $B$ is the event that at least two of $U_1,U_2,U_3$ are in the interval $({\large{\frac{1}{3}}},1]$.$\\[6pt]$
  • $C$ is the event that the median element of $U_1,U_2,U_3$ is in the interval $({\large{\frac{1}{3}}},{\large{\frac{2}{3}}})$.$\\[4pt]$

Our goal is to evaluate $P(C)$.

Noting that $C=A\cap B$, we get $$P(C)=P(A\cap B) = P(A)+P(B) - P(A\cup B)$$ Applying the binomial probability formula, we get $$P(A)=P(B) = {\small{\binom{3}{2}}}\left({\small{\frac{2}{3}}}\right)^2\left({\small{\frac{1}{3}}}\right)^1 + {\small{\binom{3}{3}}}\left({\small{\frac{2}{3}}}\right)^3\left({\small{\frac{1}{3}}}\right)^0 ={\small{\frac{20}{27}}}$$ Logically, at least one of the events $A,B$ must occur, hence $P(A\cup B)=1$, so $$P(C)={\small{\frac{20}{27}}}+{\small{\frac{20}{27}}}-1={\small{\frac{13}{27}}}$$

quasi
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We want the distribution of an order statistic. Here in our problem the order statistic is $X_2$ in $X_3>X_2>X_3$. Thus for the general case, the pdf is

$$f_{X_j}(x) = \frac{n!}{(j-1)!(n-j)!} x^{j-1}(1-x)^{n-j}.$$

In our problem we have $n=3$, and we want the pdf of the second highest statistic, so $j=2$. Hence we must consider

$$f_{X_2}(x) = 3!x(1-x) = 6x(1-x).$$

Now evaluate the pdf for the interval $(1/3,2/3)$.

Another way is to evaluate the integral

$$P = 3!.\int_{\frac{1}{3}}^{\frac{2}{3}}\int_{0}^{x_2}\int_{x_2}^{1} dx_3 dx_1 dx_2.$$

Both methods will yield $\frac{13}{27}$.

Mars Plastic
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Satish Ramanathan
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