Argue on elementary grounds that the average value of $x/y$ exceeds 1 if x and y are chosen randomly between 1 and 2

Question

Textbook problem:

Two numbers, first $x$ and then $y$, are chosen at random between $1$ and $2$. What is the average value of the quotient $\frac{x}{y}$? Can you argue on elementary grounds that the answer must exceed 1?

My answer: If $x$ has been picked, then the average of the quotient is $$ \int_1^2 \frac{x}{y}\,dy = x\ln(2) $$ Averaging over possible values of $x$ yields $$ \int_1^2 x\ln(2)\, dx = \frac{3}{2}\ln(2) \approx 1.03 $$ But this is not an argument based on elementary grounds I don't think.

Interpreting the average value of a function over an interval as being the height of the rectangle with base on the interval and with the same area as the area under the function over the interval leads me to the following argument.

The average value of $x$ is $3/2$. The quotient $x/y$ then ranges from $3/4$ to $3/2$. We can underestimate the area under this curve by splitting it up into a rectangle with length $1$ and height $3/4$ and a right triangle on top of the rectangle whose hypotenuse is the tangent line to the curve $\frac{3}{2y}$ at the point $(3/2,1)$. This gives a triangle with height $7/12$ and base $7/8$.

So an underestimate for the area under the curve would be $\left(1\cdot \frac{3}{4}\right) + \left(\frac{1}{2}\cdot \frac{7}{8}\cdot \frac{7}{12}\right) = \frac{193}{192}$.

Here is an illustration of my quotient function with $x = 3/2$, the tangent line, and the rectangle:

Since the base of the rectangle with the same area on $[1,2]$ has length $1$ its height must be at least $\frac{193}{192}$ to match the area under the graph. Hence, the height, or average value, exceeds $1$.

Question: Does this seem to be an argument the author could be looking for instead of integrating?

This is a single variable calculus text. The section is ''The Average Value of a Function". Note that not until the next chapter is probability introduced.

Update: I have obtained a copy of the author's own solutions manual. Here is the official solution:

To argue on elementary grounds that the answer must exceed 1, note that the outcomes $a/b$ and $b/a$ must occur equally often, so that the answer is certainly greater than the minimum over all $a$ and $b$ in $[1,2]$ of the average of these two numbers. Now this average is $$ \frac{1}{2}\left[\frac{a}{b} + \frac{b}{a}\right] = \frac{a^2 + b^2}{2ab} $$ and this is always greater than 1 because $a^2 + b^2 - 2ab = (a-b)^2 > 0$.

Answer 1

We can argue as follows. The probability distribution is symmetric w.r.t. interchanging $x$ and $y$. This means that we can compute the expectation value by restricting $x$ to be larger than or equal to $y$ and computing the expectation value of $f(x,y) = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}\right)$ over this modified probability distribution.

The fact that the expectation value exceeds $1$ then follows from the fact that the function $f(x,y)$ is larger than or equal to $1$:

$$f(x,y) - 1 = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}-2\right) = \frac{x^2+y^2-2xy}{2xy} = \frac{(x-y)^2}{2 xy}\geq 0$$

The function $f(x,y)$ is then equal to $1$ when $x = y$ which occurs on a subset of measure zero, so the expectation value is clearly larger than $1$.

Answer 2

The answer is in fact $\frac{3}{2} \log 2$, as correctly stated in the question.

The flaw in the geometric argument is that the average being taken is an arithmetic mean, not a geometric mean. To illustrate, suppose that instead of $X$ and $Y$ being chosen on $[1,2]$, consider instead the discrete case where $X, Y \in \{1, 2\}$. Then we have for instance $\Pr[X/Y = 2] = \Pr[X/Y = 1/2]$ but the expectation is $\frac{\frac{1}{2} + 2}{2} > 1$.

This reasoning suggests that a modification to the geometric argument will suffice in proving the expectation exceeds $1$. All that is needed is an application of the AM-GM inequality and a limiting argument.

Suppose $X$ and $Y$ are jointly discrete uniform on the set $$\left\{1 + \frac{k}{n}\right\}_{k = 1}^n.$$ Then the above argument shows that the conditional expectation of $Z = X/Y$ given $Z \in \{z, 1/z\}$ is $$\frac{z + 1/z}{2} \ge \sqrt{z \cdot 1/z} = 1.$$ Then take again the expectation over all such $z$ in the support, and we see that the total expectation must exceed $1$. Then in the limit as $n \to \infty$, $X$ and $Y$ become continuous uniform on $[1,2]$ and we are done.

A less formal way to reason, then, is to observe that for any point in the square $[1,2] \times [1,2]$, there is a matching point whose slope is the reciprocal; their average slope, by the AM-GM inequality, must exceed $1$, so the average taken over all such points in the square will also exceed $1$.

Answer 3

A calculus-free approach:

Consider the sequence $p_n, n > 0$ given by $p_n = \frac{2n}{n+1}$. Draw line segments from the origin to the points $(p_n, 2)$, and to the points $(2, p_n)$. Where these segments cross the square, they divide it into wedges. Call the wedges in the upper half of the square $A_1, A_2, A_3, \ldots$, starting from the upper left corner, and call the wedges in the lower half of the square $B_1, B_2, B_3, \ldots$, starting from the lower right corner.

Notice that the ratio $x/y$ increases monotonically as we proceed from left to right. Now, let $a_n$ be the average value of the ratio $x/y$ in the wedge $A_n$, and similarly let $b_n$ be the average value of the ratio $x/y$ in the wedge $B_n$. Because $x/y$ increases from left to right, we have that $a_n > \frac{p_n}{2} = \frac{n}{n+1}$ and $b_n > \frac{2}{p_{n+1}} = \frac{n+2}{n+1}$. Since $A_n$ and $B_n$ have equal areas, the average of the ratio $x/y$ across both areas is $\frac{a_n+b_n}{2} > 1$.

Finally, since the wedges $A_n$ and $B_n$ collectively cover the entire square, the average of the ratio $x/y$ over the entire square is greater than $1$.