Monotone likelihood ratio

Suppose you are working on a project, and you can either work hard or slack off. Call your choice of effort ${\displaystyle e}$ and the quality of the resulting project ${\displaystyle q}$. If the MLRP holds for the distribution of q conditional on your effort ${\displaystyle e}$, the higher the quality the more likely you worked hard. Conversely, the lower the quality the more likely you slacked off.

1. Choose effort ${\displaystyle e\in \{H,L\}}$ where H means high, L means low
2. Observe ${\displaystyle q}$ drawn from ${\displaystyle f(q\mid e)}$. By Bayes' law with a uniform prior,

   ${\displaystyle \Pr[e=H\mid q]={\frac {f(q\mid H)}{f(q\mid H)+f(q\mid L)}}}$

3. Suppose ${\displaystyle f(q\mid e)}$ satisfies the MLRP. Rearranging, the probability the worker worked hard is

${\displaystyle {\frac {1}{1+f(q\mid L)/f(q\mid H)}}}$

which, thanks to the MLRP, is monotonically increasing in ${\displaystyle q}$ (because ${\displaystyle f(q\mid L)/f(q\mid H)}$ is decreasing in ${\displaystyle q}$). Hence if some employer is doing a "performance review" he can infer his employee's behavior from the merits of his work.

If the family of random variables has the MLRP in ${\displaystyle T(X)}$, a uniformly most powerful test can easily be determined for the hypothesis ${\displaystyle H_{0}:\theta \leq \theta _{0}}$ versus ${\displaystyle H_{1}:\theta >\theta _{0}}$.

Example: Let ${\displaystyle e}$ be an input into a stochastic technology – worker's effort, for instance – and ${\displaystyle y}$ its output, the likelihood of which is described by a probability density function ${\displaystyle f(y;e).}$ Then the monotone likelihood ratio property (MLRP) of the family ${\displaystyle f}$ is expressed as follows: for any ${\displaystyle e_{1},e_{2}}$, the fact that ${\displaystyle e_{2}>e_{1}}$ implies that the ratio ${\displaystyle f(y;e_{2})/f(y;e_{1})}$ is increasing in ${\displaystyle y}$.

One-parameter exponential families have monotone likelihood-functions. In particular, the one-dimensional exponential family of probability density functions or probability mass functions with

${\displaystyle f_{\theta }(x)=c(\theta )h(x)\exp(\pi (\theta )T(x))}$

has a monotone non-decreasing likelihood ratio in the sufficient statistic T(x), provided that ${\displaystyle \pi (\theta )}$ is non-decreasing.

Monotone likelihood functions are used to construct uniformly most powerful tests, according to the Karlin–Rubin theorem.[1] Consider a scalar measurement having a probability density function parameterized by a scalar parameter θ, and define the likelihood ratio ${\displaystyle \ell (x)=f_{\theta _{1}}(x)/f_{\theta _{0}}(x)}$. If ${\displaystyle \ell (x)}$ is monotone non-decreasing, in ${\displaystyle x}$, for any pair ${\displaystyle \theta _{1}\geq \theta _{0}}$ (meaning that the greater ${\displaystyle x}$ is, the more likely ${\displaystyle H_{1}}$ is), then the threshold test:

${\displaystyle \varphi (x)={\begin{cases}1&{\text{if }}x>x_{0}\\0&{\text{if }}x<x_{0}\end{cases}}}$

where ${\displaystyle x_{0}}$ is chosen so that ${\displaystyle \operatorname {E} _{\theta _{0}}\varphi (X)=\alpha }$

is the UMP test of size α for testing ${\displaystyle H_{0}:\theta \leq \theta _{0}{\text{ vs. }}H_{1}:\theta >\theta _{0}.}$

Note that exactly the same test is also UMP for testing ${\displaystyle H_{0}:\theta =\theta _{0}{\text{ vs. }}H_{1}:\theta >\theta _{0}.}$

Monotone likelihood-functions are used to construct median-unbiased estimators, using methods specified by Johann Pfanzagl and others.[2][3] One such procedure is an analogue of the Rao–Blackwell procedure for mean-unbiased estimators: The procedure holds for a smaller class of probability distributions than does the Rao–Blackwell procedure for mean-unbiased estimation but for a larger class of loss functions.[3]^(p713)

If a family of distributions ${\displaystyle f_{\theta }(x)}$ has the monotone likelihood ratio property in ${\displaystyle T(X)}$,

1. the family has monotone decreasing hazard rates in ${\displaystyle \theta }$ (but not necessarily in ${\displaystyle T(X)}$)
2. the family exhibits the first-order (and hence second-order) stochastic dominance in ${\displaystyle x}$, and the best Bayesian update of ${\displaystyle \theta }$ is increasing in ${\displaystyle T(X)}$.

But not conversely: neither monotone hazard rates nor stochastic dominance imply the MLRP.

Let distribution family ${\displaystyle f_{\theta }}$ satisfy MLR in x, so that for ${\displaystyle \theta _{1}>\theta _{0}}$ and ${\displaystyle x_{1}>x_{0}}$:

${\displaystyle {\frac {f_{\theta _{1}}(x_{1})}{f_{\theta _{0}}(x_{1})}}\geq {\frac {f_{\theta _{1}}(x_{0})}{f_{\theta _{0}}(x_{0})}},}$

or equivalently:

${\displaystyle f_{\theta _{1}}(x_{1})f_{\theta _{0}}(x_{0})\geq f_{\theta _{1}}(x_{0})f_{\theta _{0}}(x_{1}).\,}$

Integrating this expression twice, we obtain:

1. To ${\displaystyle x_{1}}$ with respect to ${\displaystyle x_{0}}$ ${\displaystyle {\begin{aligned}&\int _{\min _{x}\in X}^{x_{1}}f_{\theta _{1}}(x_{1})f_{\theta _{0}}(x_{0})\,dx_{0}\\[6pt]\geq {}&\int _{\min _{x}\in X}^{x_{1}}f_{\theta _{1}}(x_{0})f_{\theta _{0}}(x_{1})\,dx_{0}\end{aligned}}}$ integrate and rearrange to obtain ${\displaystyle {\frac {f_{\theta _{1}}}{f_{\theta _{0}}}}(x)\geq {\frac {F_{\theta _{1}}}{F_{\theta _{0}}}}(x)}$		2. From ${\displaystyle x_{0}}$ with respect to ${\displaystyle x_{1}}$ ${\displaystyle {\begin{aligned}&\int _{x_{0}}^{\max _{x}\in X}f_{\theta _{1}}(x_{1})f_{\theta _{0}}(x_{0})\,dx_{1}\\[6pt]\geq {}&\int _{x_{0}}^{\max _{x}\in X}f_{\theta _{1}}(x_{0})f_{\theta _{0}}(x_{1})\,dx_{1}\end{aligned}}}$ integrate and rearrange to obtain ${\displaystyle {\frac {1-F_{\theta _{1}}(x)}{1-F_{\theta _{0}}(x)}}\geq {\frac {f_{\theta _{1}}}{f_{\theta _{0}}}}(x)}$

Combine the two inequalities above to get first-order dominance:

${\displaystyle F_{\theta _{1}}(x)\leq F_{\theta _{0}}(x)\ \forall x}$

Use only the second inequality above to get a monotone hazard rate:

${\displaystyle {\frac {f_{\theta _{1}}(x)}{1-F_{\theta _{1}}(x)}}\leq {\frac {f_{\theta _{0}}(x)}{1-F_{\theta _{0}}(x)}}\ \forall x}$

The MLR is an important condition on the type distribution of agents in mechanism design. Most solutions to mechanism design models assume a type distribution to satisfy the MLR to take advantage of a common solution method.