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## MHR, Regular Distributions and Myerson’s Lemma

Monotone Hazard Rate (MHR) distributions and its superclass regular distributions keep appearing in the Mechanism Design literature and this is due to a very good reason: they are the class of distributions for which Myerson’s Optimal Auction is simple and natural. Let’s brief discuss some properties of those distributions. First, two definitions:

1. Hazard rate of a distribution $f$ : $h(z) = \frac{f(z)}{1-F(z)}$
2. Myerson virtual value of a distribution $f$ : $\phi(z) = z - \frac{1-F(z)}{f(z)}$

We can interpret the hazard rate in the following way: think of $T \sim f$ as a random variable that indicates the time that a light bulb will take to extinguish. If we are in time $t$ and the light bulb hasn’t extinguished so far, what is the probability it will extinguish in the next $\delta$ time:

$\mathbb{P}[T \leq t+\delta \vert T > t] \approx \frac{f(t) \delta}{1-F(t)}$

We say that a distribution is monotone hazard rate, if $h(z)$ is non-decreasing. This is very natural for light bulbs, for example. Many of the distributions that we are used to are MHR, for example, uniform, exponential and normal. The way that I like to think about MHR distributions is the following: if some distribution has hazard rate $h(z)$, then it means that $F'(z) = (1-F(z)) h(z)$. If we define $G(z) = 1-F(z)$, then $(log G(z))' = \frac{G'(z)}{G(z)} = -h(z)$, so:

$F(z) = 1-\text{exp}(-\int_0^z h(u) du)$

From this characterization, it is simple to see that the extremal distributions for this class, i.e. the distributions that are in the edge of being MHR and non-MHR are constant hazard rate, which correspond to the exponential distribution $F(z) = 1-e^{-\lambda z}$ for $z \in [0,\infty)$. They way I like to think about those distributions is that whenever you are able to prove something about the exponential distribution, then you can prove a similar statement about MHR distributions. Consider those three examples:

Example 1: $\mathbb{P}[\phi(z) \geq 0] \geq \frac{1}{e}$ for MHR distributions. This fact is straightforward for the exponential distribution. For the exponential distribution $\phi(z) = z-\lambda^{-1}$ and therefore

$\mathbb{P}[\phi(z) \geq 0] \geq \mathbb{P}[z > \lambda^{-1}] = 1-F(\lambda^{-1}) = e^{-1}$

but the proof for MHR is equally simple: Let $r = \inf \{z; \phi(z) \geq 0\}$, therefore $r h(r) \leq 1$.
$P\{ \phi(v) \geq 0\} = P\{ v \geq r \} = 1 - F(r) = e^{-\int_0^r h(u) du} \geq e^{-r h(r)} \geq e^{-1}$

Example 2: Given $z_1, z_2 \sim f$ iid where $f$ is MHR and $v_1 = \max \{z_1, z_2\}$ and $v_2 = \min \{z_1, z_2\}$, then $\mathbb{E}[v_2] \geq \frac{1}{3} \mathbb{E}[v_1]$. The proof for the exponential distribution is trivial, and in fact, this is tight for the exponential, the trick is to use the convexity of $z \mapsto \int_0^z h(u) du$. We use that $\int_0^{2z} h \geq 2 \int_0^z h$ in the following way:

$\mathbb{E} [v_2] = \int_0^\infty (1 - F(z))^2 dz = \int_0^\infty e^{-2 \int_0^z h} dz$

$\geq \int_0^\infty e^{-\int_0^{2z} h} dz= \frac{1}{2} \int_0^\infty 1 - F(z) dz = \frac{1}{2} \mathbb{E} [z]$

Since $\mathbb{E} [v_1 + v_2] = \mathbb{E} [z_1 + z_2] = 2 \mathbb{E} [z]$, we have that $\mathbb{E}[v_1] = 2 \mathbb{E}[z] - \mathbb{E}[v_2] \leq \frac{3}{2} \mathbb{E}[z]$. This way, we get: $\mathbb{E}[v_2] \geq \frac{1}{2}\mathbb{E}[z] \geq \frac{1}{2} \cdot \frac{2}{3} \mathbb{E}[v_1] = \frac{1}{3} \mathbb{E}[v_1]$

Example 3: For MHR distributions, there is a simple lemma that relates the virtual value and the real value and this lemma is quite useful in various settings: let $r = \inf \{z; \phi(z) > 0 \}$, then for $z \geq r$, $\phi(z) \geq z - r$. Again, this is tight for exponential distribution. The proof is quite trivial:

$x - \phi(x) = \frac{1-F(x)}{f(x)} \leq \frac{1-F(r)}{f(r)} = r$

Now, MHR distributions are a subclass of regular distributions, which are the distributions for which Myerson’s virtual value $\phi(z)$ is a monotone function. I usually find harder to think about regular distributions than to think about MHR (in fact, I don’t know so many examples that are regular, but not MHR. Here is one, though, called the equal-revenue-distribution. Consider $z \in [1, \infty)$ distributed according to $f(z) = 1/z^2$. The cumulative distribution is given by $F(z) = 1-1/z$. The interesting thing of this distribution is that posted prices get the same revenue regardless of the price. For example, if we post any price $r \in [1,\infty)$, then a customer with valuations $z \sim f$ buys the item if $z > r$ by price $r$, getting  revenue is $r (1-F(r)) = 1$. This can be expressed by the fact that $\phi(z) = 0$. I was a bit puzzled by this fact, because of Myerson’s Lemma:

Myerson Lemma: If a mechanism sells to some player that has valuation $v \sim f$ with probability $x(v)$ when he has value $v$, then the revenue is $\mathbb{E} [x(v) \phi(v)]$.

And it seemed that the auctioneers was doomed to get zero revenue, since $\phi(z) = 0$. For example, suppose we fix some price $r$ and we sell the item if $v \geq r$ by price $r$. Then it seems that Myerson’s Lemma should go through by a derivation like that (for this special case, although the general proof is quite similar):

$\mathbb{E} [x(v) \phi(v)] = \int_r^\infty \phi(z) f(z) dz = \int_r^\infty z f(z) - (1-F(z)) dz =$

$= \int_r^\infty [ z f(z) - \int_z^\infty f(u) du ] dz = \int_r^\infty z f(z) dz - \int_r^\infty \int_r^u f(u) dz du$

$= r (1-F(r))$

but those don’t seem to match, since one side is zero and the other is 1. The mistake we did above is classic, which is to calculate $\infty - \infty$. We wrote:

$\mathbb{E}[\phi(v)] = \int_r^\infty z f(z) dz - \int_r^\infty 1-F(z) dz$

but both are infinity! This made me realize that Myerson’s Lemma needs the condition that $\mathbb{E}[z] < \infty$, which is quite a natural a distribution over valuations of a good. So, one of the bugs of the the equal-revenue-distribution is that $\mathbb{E}[z] = \infty$. A family that is close to this, but doesn’t suffer this bug is: $f(z) = \frac{\alpha-1}{z^\alpha}$ for $z \in [1,\infty)$, then $F(z) = 1 - z^{1-\alpha}$. For $\alpha > 2$ we have $\mathbb{E}[v] < \infty$, then we get $\phi(z) = \frac{\alpha-2}{\alpha-1} z$.

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