The expected value of the lower of two values in a uniforme distribuition in [0,1], one of which is known to exceed $\frac{1}{2}$

Question

I am reading a paper about Auction Theory, named "Auctions Versus Negotiations" by Jeremy Bulow and Paul Klemperer - 1996. Anyway, I just saw two affirmations about probabilities that I don't know exactly how to prove.

$\underline{\textbf{First}}$: The expected value of the lower of two values in a uniform distribution in $[0,1]$ is $\frac{1}{3}$.

This one seems like is I divide the interval in quantiles I would find the result. $(0, \frac{1}{3}, \frac{2}{3}, 1)$

$\underline{\textbf{Second}}$: The expected value of the lower of two values in a uniform distribution in $[0,1]$, one of which is known to exceed $\frac{1}{2}$, is $\frac{11}{24}$.

Since the expected value of an uniform distribution in $[0,1]$ is:

$$ \int_0^1 \frac{x}{1-0} \; dx $$

I tried to solve this integral:

$$ \int_0^x \frac{1}{x} y \; dy, \text{ knowing that } x> \frac{1}{2} $$

As result I got:

$$ \frac{x}{2}, \text{ knowing that } x> \frac{1}{2}$$

But it doesn't look a good direction. I'd appreciated if someone could give me any help.

First of all I would like to thank you for having read this far.

Answer 1

This answer assumes that the two values are independent (if they aren't, it's impossible to solve the problem without knowing their joint distribution).

First question: this can be calculated with the integral $$ \int_0^1 \int_0^1 \min(x,y) \,dx\,dy = \int_0^1 \int_0^x y\,dy\,dx + \int_0^1 \int_x^1 x \,dy\,dx = \frac13. $$

Second question: the same general idea applies, but it's a conditional expectation. So instead we have to calculate $$ \frac{\iint_A \min(x,y) \,dx\,dy}{\mathbb{P}(A)} $$ where $A$ is the set $\{(x,y) : x>1/2 \ \ \text{or}\ \ y > 1/2 \}$. This is evaluated to \begin{align*} &\frac{1}{3/4} \left( \int_0^{1/2} \int_{1/2}^1 x \,dy\,dx + \int_{1/2}^1 \int_0^x y \,dy\,dx + \int_{1/2}^1 \int_x^1 x\,dy\,dx \right) \\ &= \ \frac{7}{18}. \end{align*}

(Hint: if you don't see where these integrals come from, try drawing a picture)

As antkam points out in the comments, conditioning on "one of the numbers being >1/2" is different than conditioning on "the first number being >1/2", which is why 11/24 is not actually the correct answer.

Answer 2

The joint pdf of the two values is $f(x,y)=1$. So we can calculate the expectation of the lower as follows:

$$\int_0^1\int_0^1f(x,y)\min(x,y)\,dy\,dx=\int_0^1\left(\int_0^xy\,dy+\int_x^1x\,dy\right)\,dx\\ =\int_0^1(x^2/2+x(1-x))\,dx\\ =\left[x^2/2-x^3/6\right]_0^1=1/3.$$

When we condition on the larger of the two being greater than $1/2$, we change the pdf: it becomes $0$ whenever this is not true, and is scaled up appropriately in the other region. So we have a new pdf $$g(x,y)=\begin{cases}4/3\text{ if }\max(x,y)>1/2\\0\text{ otherwise.}\end{cases}$$ We could in principle do the same thing with this new pdf to get the expectation. However, there is a simpler way. There are three cases remaining, each equally likely. Either $x<1/2<y$, or $y<1/2<x$, or $1/2<x,y$. In either of the first two cases the distribution of the minimum is $\mathrm{Unif}(0,1/2)$, whereas in the last case it is the minimum of two independent $\mathrm{Unif}(1/2,1)$ random variables, and we can use the above result to deduce its expectation.

(I get an answer of $7/18$ for this interpretation of the problem.)