Comparing Distributions by Their Ratios
Likelihood-ratio order turns a distributional comparison into a question about how relative probability changes across the possible values. If one law puts increasingly more weight on larger outcomes relative to another, its mass or density ratio rises with the outcome. That makes the order a particularly direct tool for comparing members of familiar parametric families.
The direction of the ratio matters. In this tutorial, \(X\le_{\mathrm{lr}}Y\) means that the law of \(Y\), in the numerator, is larger in likelihood-ratio order than the law of \(X\). We compare Bernoulli laws, binomial laws with the same number of trials, Poisson laws, and exponential laws. The parameter directions are not all the same: in particular, increasing an exponential rate shifts the distribution in the opposite direction from increasing its mean.
For example, if both mass functions are positive at \(x<y\), the cross-product inequality can be divided by \(f_X(x)f_X(y)\). It becomes \(f_Y(y)/f_X(y)\ge f_Y(x)/f_X(x)\). If one mass is zero, division may be impossible, so the cross-product definition is the safer general statement. This is the support issue emphasized in likelihood-ratio comparisons.
Bernoulli and Binomial Laws
Let \(X\sim\operatorname{Bernoulli}(p)\) and \(Y\sim\operatorname{Bernoulli}(q)\), where the value is either 0 or 1. For interior parameters \(0<p,q<1\), the only nontrivial comparison of support points is 0 with 1. The cross-product condition is $$ f_Y(1)f_X(0)\ge f_Y(0)f_X(1) \quad\Longleftrightarrow\quad q(1-p)\ge (1-q)p \quad\Longleftrightarrow\quad q\ge p. $$ Equivalently, \(f_Y(1)/f_X(1)\) is at least \(f_Y(0)/f_X(0)\) precisely when the success probability \(q\) is at least \(p\).
Worked Example: Bernoulli Success Probabilities
Compare \(X\sim\operatorname{Bernoulli}(0.25)\) and \(Y\sim\operatorname{Bernoulli}(0.60)\). Their mass ratios at 0 and 1 are $$ \frac{f_Y(0)}{f_X(0)} =\frac{0.40}{0.75} =\frac{8}{15}, \qquad \frac{f_Y(1)}{f_X(1)} =\frac{0.60}{0.25} =\frac{12}{5}. $$ Since \(8/15<12/5\), the ratio increases from 0 to 1 and \(X\le_{\mathrm{lr}}Y\). Directly, the cross products are \(0.60(0.75)=0.45\) and \(0.40(0.25)=0.10\), so the required inequality holds.
Now fix the number of trials \(n\) and compare \(X\sim\operatorname{Binomial}(n,p)\) with \(Y\sim\operatorname{Binomial}(n,q)\). For \(0<p,q<1\), the ratio at \(k\in\{0,\ldots,n\}\) is $$ \frac{f_Y(k)}{f_X(k)} =\left(\frac{1-q}{1-p}\right)^n \left(\frac{q(1-p)}{p(1-q)}\right)^k. $$ As \(k\) increases by one, this ratio is multiplied by \(q(1-p)/[p(1-q)]\). That multiplier is at least one exactly when \(q\ge p\). Thus increasing the success probability gives likelihood-ratio order for a binomial distribution with fixed \(n\).
Worked Example: Binomial Success Probability
Compare \(X\sim\operatorname{Binomial}(5,0.30)\) and \(Y\sim\operatorname{Binomial}(5,0.60)\). For every \(k=0,\ldots,5\), $$ \frac{f_Y(k)}{f_X(k)} =\left(\frac{0.40}{0.70}\right)^5 \left(\frac{0.60(0.70)}{0.30(0.40)}\right)^k =\left(\frac{4}{7}\right)^5\left(\frac{7}{2}\right)^k. $$ The ratio is multiplied by \(7/2>1\) at each successive value of \(k\). For a numerical check, it is \((4/7)^5=1024/16807\) at \(k=0\), and \(32\) at \(k=5\), since \((4/7)^5(7/2)^5=2^5\). The ratio is increasing throughout the support, so \(X\le_{\mathrm{lr}}Y\).
The endpoint probabilities \(p=0\) and \(p=1\) give degenerate binomial laws, so the interior ratio formula does not apply there. The cross-product definition still does. If \(p=0\le q\), \(X\) is concentrated at 0 and the comparison holds: for \(0<y\le n\), \(f_Y(y)f_X(0)\ge0=f_Y(0)f_X(y)\). If \(q=1\), \(Y\) is concentrated at \(n\); for \(x<y<n\), both \(f_Y(x)\) and \(f_Y(y)\) are zero, while for \(y=n\) the cross-product inequality is nonnegative. Thus the same direction holds at the endpoints as well.
Poisson Laws
For \(X\sim\operatorname{Poisson}(\lambda_1)\) and \(Y\sim\operatorname{Poisson}(\lambda_2)\), first suppose both parameters are positive. At \(k=0,1,\ldots\), $$ \frac{f_Y(k)}{f_X(k)} =e^{-(\lambda_2-\lambda_1)} \left(\frac{\lambda_2}{\lambda_1}\right)^k. $$ The ratio is multiplied by \(\lambda_2/\lambda_1\) when \(k\) increases by one. It is therefore nondecreasing exactly when \(\lambda_2\ge\lambda_1\): a larger Poisson mean gives the larger law in likelihood-ratio order.
Worked Example: Poisson Means
Compare \(X\sim\operatorname{Poisson}(2)\) and \(Y\sim\operatorname{Poisson}(5)\). The ratio at \(k\) is $$ \frac{f_Y(k)}{f_X(k)} =e^{-3}\left(\frac{5}{2}\right)^k. $$ Each step to the next integer multiplies the ratio by \(5/2>1\). For instance, the ratio is \(e^{-3}\) at \(k=0\) and \((25/4)e^{-3}\) at \(k=2\). Hence \(X\le_{\mathrm{lr}}Y\). The calculation compares the full mass functions; the factorial term \(1/k!\) cancels in their ratio.
If \(\lambda_1=0\), the first law is concentrated at 0. For \(\lambda_2>0\), the cross-product condition holds: for any \(y>0\), \(f_Y(y)f_X(0)\ge0=f_Y(0)f_X(y)\), and pairs with \(x>0\) have \(f_X(x)=0\). If \(\lambda_2=0\), both laws are the same point mass when \(\lambda_1=0\); when \(\lambda_1>0\), the order \(X\le_{\mathrm{lr}}Y\) does not hold. For example, taking \(x=0<y=1\) gives \(f_Y(1)f_X(0)=0<f_Y(0)f_X(1)\). These boundary checks agree with the parameter direction \(\lambda_1\le\lambda_2\).
Exponential Laws: Rate Reverses the Direction
Use the rate parameterization: an exponential random variable with rate \(\lambda>0\) has density \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge0\), and density zero for \(x<0\). Compare \(X\) with rate \(\lambda_1\) to \(Y\) with rate \(\lambda_2\). On their common support, $$ \frac{f_Y(x)}{f_X(x)} =\frac{\lambda_2}{\lambda_1} e^{-(\lambda_2-\lambda_1)x}. $$ This ratio is nondecreasing in \(x\) exactly when \(\lambda_2\le\lambda_1\). Thus a smaller rate, corresponding to a longer typical lifetime, gives the larger exponential law in likelihood-ratio order.
Worked Example: Exponential Rates
Let \(X\) have exponential rate 3 and \(Y\) have exponential rate 1.5. Their density ratio for \(x\ge0\) is $$ \frac{f_Y(x)}{f_X(x)} =\frac{1.5}{3}e^{-(1.5-3)x} =\frac12 e^{1.5x}. $$ Its derivative is \(\frac34e^{1.5x}>0\), so it is strictly increasing. For example, the ratio is \(1/2\) at \(x=0\), and \(\frac12e^3\) at \(x=2\), which is larger. Therefore \(X\le_{\mathrm{lr}}Y\). The comparison is consistent with the means \(1/3\) and \(1/1.5=2/3\): \(Y\) has the longer mean lifetime.
Ratio Criterion and Parameter Directions
To prove the theorem, take any two support points \(x<y\). Positivity permits division by \(f_X(x)f_X(y)>0\). The cross-product condition is equivalent to $$ f_Y(y)f_X(x)\ge f_Y(x)f_X(y) \quad\Longleftrightarrow\quad \frac{f_Y(y)}{f_X(y)}\ge\frac{f_Y(x)}{f_X(x)} \quad\Longleftrightarrow\quad r(y)\ge r(x). $$ If the cross-product condition holds for every ordered pair, then \(r\) is nondecreasing. Conversely, if \(r\) is nondecreasing, the same equivalence gives the cross-product condition for every pair. This proves both directions. \(\square\)
The family calculations above use this result on supports where the compared mass functions or densities are positive. At boundaries where positivity fails, the cross-product definition must be checked directly, as for the degenerate Bernoulli, binomial, and Poisson cases. The resulting directions are summarized here.
| Family and comparison | Condition for \(X\le_{\mathrm{lr}}Y\) | Ratio behavior |
|---|---|---|
| Bernoulli: parameters \(p\) and \(q\) | \(p\le q\) | Increasing from 0 to 1 |
| Binomial: same \(n\), parameters \(p\) and \(q\) | \(p\le q\) | Successive ratio multiplier is at least 1 |
| Poisson: means \(\lambda_1\) and \(\lambda_2\) | \(\lambda_1\le\lambda_2\) | Successive ratio multiplier is at least 1 |
| Exponential: rates \(\lambda_1\) and \(\lambda_2\) | \(\lambda_1\ge\lambda_2\) | Density ratio is nondecreasing in \(x\) |
Common Mistake and Why the Order Matters
A frequent error is to reverse the ratio without reversing the order. If the goal is to check \(X\le_{\mathrm{lr}}Y\), use \(f_Y/f_X\), not \(f_X/f_Y\). For the exponential example, \(f_Y/f_X=\frac12e^{1.5x}\) increases, whereas its reciprocal decreases. The reciprocal’s behavior does not contradict the comparison; it answers the question with the two distributions swapped.
A second trap is to say that a higher exponential rate gives a larger law merely because the rate parameter is larger. The rate is inversely related to the mean. The ratio calculation shows exactly why the direction is reversed relative to the Bernoulli success probability, the binomial success probability, and the Poisson mean. Always identify the parameterization and derive the ratio rather than relying on a parameter’s name.
Likelihood-ratio order is a strong comparison: it tracks the relative weighting of outcomes across the whole support, not just a difference in means. The earlier material on the implication from likelihood-ratio order to hazard-rate and usual stochastic order explains one consequence of having this stronger order. Here the practical lesson is to simplify the ratio, check its monotonicity, and then verify any support boundaries before concluding that the comparison holds.
Check Your Understanding
Use the numerator convention \(f_Y/f_X\), and check support conditions when a parameter is at a boundary.
- For \(X\sim\operatorname{Bernoulli}(0.4)\) and \(Y\sim\operatorname{Bernoulli}(0.7)\), calculate the mass ratio at 0 and at 1. Which law is larger in likelihood-ratio order?
- For binomial laws with the same \(n\), why is the successive ratio multiplier more useful than comparing the ratios only at the two endpoints of the support?
- Compare \(\operatorname{Poisson}(6)\) with \(\operatorname{Poisson}(4)\). What multiplier relates successive values of the ratio \(f_Y(k)/f_X(k)\), and what does it imply?
- Let \(X\) have exponential rate 2 and \(Y\) have exponential rate 4. Determine the direction of likelihood-ratio order by writing the density ratio.
- Why is the cross-product definition needed when one of two compared laws is degenerate?