A/B Testing Measurement Framework for Recommendation Models Based on Expected Revenue

TL;DR

This paper introduces a statistical framework for evaluating whether new recommendation systems improve revenue per visit by analyzing conversion rate and order value, using tests that require smaller samples and provide confidence intervals.

Contribution

It presents a novel A/B testing method for recommendation models that efficiently detects revenue improvements with smaller samples and no subjective outlier removal.

Findings

Method requires fewer samples than existing approaches.

Assumes log-normal distribution for purchase values, validated empirically.

Provides confidence intervals to quantify uncertainty.

Abstract

We provide a method to determine whether a new recommendation system improves the revenue per visit (RPV) compared to the status quo. We achieve our goal by splitting RPV into conversion rate and average order value (AOV). We use the two-part test suggested by Lachenbruch to determine if the data generating process in the new system is different. In cases that this test does not give us a definitive answer about the change in RPV, we propose two alternative tests to determine if RPV has changed. Both of these tests rely on the assumption that non-zero purchase values follow a log-normal distribution. We empirically validate this assumption using data collected at different points in time from Staples.com. On average, our method needs a smaller sample size than other methods. Furthermore, it does not require any subjective outlier removal. Finally, it characterizes the uncertainty around RPV by providing a confidence interval.

Tables

Table 1. Table 1: Expected Revenue Significance Test

	$AOV$ Pos. Sig.	$AOV$ Non-Sig.	$AOV$ Neg. Sig.
$r$ Pos. Sig.	+	+	X
$r$ Non-Sig.	+	0	-
$r$ Neg. Sig.	X	-	-

Equations

$$RPV=r.AOV.$$

$$z_p=\frac{p_1-p_2}{\sqrt{p_p(1-p_p)(\frac{1}{n_1}+\frac{1}{n_2})}}$$

$$z_U=\frac{U-\frac{n_1{n}_2}{2}}{\sqrt{n_1{n}_2(n_1+n_2+1)/12}}$$

$${\sigma }_{RPV}^2={\widehat{AOV}}^2{\hat{\sigma }}_r^2+{\hat{r}}^2{\hat{\sigma }}_{AOV}^2$$

$${\hat{\sigma }}_r^2=\hat{r}(1-\hat{r})$$

$${\hat{\sigma }}_{RPV}^2=e^{\hat{\mu }+{\hat{\sigma }}_{AOV}^2/2}\hat{r}(1-\hat{r})+{\hat{r}}^2(e^{{\hat{\sigma }}_{AOV}^2}-1)e^{2\hat{\mu }+{\hat{\sigma }}_{AOV}^2}$$

$$h(n_1+n_2)=(\frac{1}{n_1}+\frac{1}{n_2}{)}^{-1}$$

$$({\widehat{RPV}}_{treatment}-{\widehat{RPV}}_{control})\pm 1.96\sqrt{\frac{{\hat{\sigma }}_{RPV}^2}{h(n_1,n_2)}}$$

$$LR=\frac{L(H_0)}{L(H_1)}$$

$$L(H_0)=LN(O{V}_c|{\mu }_c,{\sigma }_c^2)Bin(n_c,k_c|p_c)LN(O{V}_t|{\mu }_t,{\sigma }_t^2)Bin(n_t,k_t|p_t)$$

$$p_c{\mu }_c=p_t{\mu }_t,$$

$$L(H_1)=LN(O{V}_c|{\mu }_c,{\sigma }_c^2)Bin(n_c,k_c|p_c)LN(O{V}_t|{\mu }_t,{\sigma }_t^2)Bin(n_t,k_t|p_t)$$

$$\begin{array}[]{c}Log(H_{0})=\log LN(OV_{c}|\mu_{c},\sigma_{c}^{2})+\log Bin(n_{c},k_{c}|p_{c})+\log LN(OV_{t}|\mu_{t},\sigma_{t}^{2})\\ +\log Bin(n_{t},k_{t}|p_{t})-\lambda(p_{c}\mu_{c}-p_{t}\mu_{t})\end{array}$$

$$\begin{cases}\begin{array}[]{c}k_{c}(1-p_{c})-p_{c}(n_{c}-k_{c})-\lambda\mu_{c}p_{c}(1-p_{c})=0\\ k_{t}(1-p_{t})-p_{c}(n_{t}-k_{t})-\lambda\mu_{t}p_{t}(1-p_{t})=0\\ \log AOV_{c}-\mu_{c}-\lambda\mu_{c}^{2}p_{c}=0\\ \log AOV_{t}-\mu_{t}-\lambda\mu_{t}^{2}p_{t}=0\\ \begin{array}[]{c}(\log AOV_{c}-\mu_{c})^{2}=\sigma_{c}^{2}\\ (\log AOV_{t}-\mu_{t})^{2}=\sigma_{t}^{2}\\ \mu_{t}p_{t}-\mu_{c}p_{c}=0\end{array}\end{array}\end{cases}$$

Full text

A/B Testing Measurement Framework for Recommendation Models Based

on Expected Revenue

Meisam Hejazi nia, Majid Hosseini, Bryant Sih

Sparx -A Staples Digital Solution Innovation Lab, West 5th Avenue, 3rd Floor, San Mateo, California 94402

1 Abstract

We provide a method to determine whether a new recommendation system improves the revenue per visit ($RPV$) compared to the status quo. We achieve our goal by splitting $RPV$ into conversion rate and average order value ($AOV$). We use the two-part test suggested by Lachenbruch to determine if the data generating process in the new system is different. In cases that this test does not give us a definitive answer about the change in $RPV$, we proposed two alternative tests to determine if $RPV$ has changed. Both of these tests rely on the assumption that non-zero purchase values follow a log-normal distribution. We empirically validated this assumption using data collected at different points in time from Staples.com. On average, our method needs a smaller sample size than other methods. Furthermore, it does not require any subjective outlier removal. Finally, it characterises the uncertainty around $RPV$ by providing a confidence interval.

2 Introduction

Measurement is a never-ending journey. Every measurement method has its own drawbacks, so there is no one size fits all approach. Because for retail recommendations the vast majority of outcomes (usually ¿ 95%) are “no-purchase” events, the conversion rate is small (large clumping at no-purchases). Furthermore, there are always customers with large purchases, and thus, the distribution of non-zero order values is not normal. Using Normal theory tests in this case causes bias in estimates ([7]). In addition, outlier removal is a difficult task and in many cases subjective removal might bias estimates. For example, when data is skewed and non-normal (like the distribution of prices) studies show that sample size needs to increase ([13]). Or in the case of non-normality, bootstrap estimates might be inconsistent ([1, 14]). Finally, estimates that are efficient for clean data from simple distributions (e.g. normal distribution which is symmetric and unimodal with thin tail) may not be robust to contamination by outliers, and may be inefficient for more complicated distributions, raising the need for very large sample sizes for valid inference ([17]).

One performance metric of a recommendation model is Revenue Per Visit ($RPV$). The relationship between $RPV$, Average Order Value ($AOV$), and conversion rate r is as follows:

[TABLE]

To determine whether a new recommendation model has changed either of $r$ or $AOV$, we can use a powerful test proposed by [7]. This test combines statistics for detecting differences in conversion rate and $AOV$ to obtain a Chi-square valued measure that determines if there is a statistically significant difference between the two models. This test is powerful because it requires a small sample size. We will describe Lachenbruch’s test in the next section. In a general sense, comparisons can be defined in 3 exclusive states: significantly positive (Pos. Sig.), non-significant (Non-Sig.), or significantly negative (Neg. Sig.). Since the objective of a recommendation system is to increase $RPV$, we can visualize the outcomes of a significance test of $r$ and $AOV$ on $RPV$ based on Table 1, where + is a guaranteed significantly positive difference in $RPV$, − is a decrease, 0 is no difference, and X is indeterminate. For most of the cases, Lachenbruch’s test tells us with a relatively small sample size whether the new recommendation system is more effective. For the two indeterminate cases we develop two parametric tests of significance.

3 Tests of Significance

In this section, we will describe in detail the process to determine whether $RPV$ of the new recommendation is different from the old recommendation system in a statistically significant sense. To summarize this process, its stages are as follows:

1. Run a pilot study and determine the sample size for Lachenbruch’s Test.

2. Run Lachenbruch’s Test and if it results in one of the determinate cases, then stop the test with a definite conclusion (See Table 1).

3. If Lachenbruch’s Test is indeterminate, run a new pilot study to determine the sample size for either of our newly devised parametric tests (derived in the next section).

4. Run the new parametric test, and conclude if there is a statistically significant difference for $RPV$ between the two models.

4 Lachenbruch’s Two-Part Test

Lachenbruch’s Test applies to data with clumping at zero. Clumping at zero for recommendation systems can be interpreted as no purchase. This test compares the distribution of $RPV$ across control and treatment groups. In other words, the null hypothesis in this test asserts the equality of these two distributions. To run this test, [7] uses two statistical measures: $z_{p}$ and $z_{U}$. The value $z_{p}$ denotes z-statistics for proportion of no purchases, and it has a normal distribution in the limit, so its square has an asymptotic Chi-square distribution with one degree of freedom. The value $z_{U}$ is a function of Mann-Whitney-Wilcox U statistic for order values and has an asymptotic normal distribution, so its square has Chi-square distribution with one degree of freedom. As a result, Lachenbruch’s statistic $L=z_{p}^{2}+z_{U}^{2}$ has an asymptotic Chi-square distribution with two degrees of freedom. If $L$ is greater than 9.633 we can reject the null hypothesis with 95% confidence. The statistic $z_{p}$ is defined as follows:

[TABLE]

where $p_{1}$, $p_{2}$ denote the no-purchase probabilities for the treatment and control groups, respectively. The value $p_{p}$ denotes the no purchase probability for the pooled group (i.e. $p_{p}=\frac{k_{1}+k_{2}}{n_{1}+n_{2}}).$ The values $k_{1}$, $k_{2}$ denote the observed number of no purchases for the treatment and control groups, respectively. The values $n_{1}$, $n_{2}$ denote the sample sizes for the treatment and control groups, respectively.

The value $z_{U}$ is defined as follows:

[TABLE]

where $U$ is the Mann-Whitney-Wilcoxon statistic for order values.111Computing rank-sum $U$ involves counting the number of times the first value wins over any observations in the other set (the other value loses if the first is larger) with 0.5 for any ties. The sum of wins and ties construct the $U$ measure. $U$ can be calculated using most statistical packages or python (scipy.stats mannwhitneyu)..

If the Chi-square test suggests significant difference in the distributions of $RPV$ between the treatment and control groups, [7] suggests we can compare the values of $z_{p}$ and $z_{U}$ against the standard normal distribution (i.e. 1.96 and -1.96 as a threshold for 95% confidence) to identify the sources of the difference. Knowing the sources of the difference between the two distributions, we can use Table 1 in order to identify whether the expected revenue between treatment and control groups are statistically different. In the cases where Table 1 is indeterminate, we need to use another statistical test. This test requires collecting a new data set (i.e. we can not use the old data set for the new test).

5 Tests for Detecting Change In Expected Revenue

In this section, we introduce two tests that assume a parametric form for $RPV$ and attempt to establish if there is any change in $RPV$ between treatment and control groups. Both tests assume that the data is generated in a two-stage process, which mixes non-purchases with log-normally distributed order values (See Figure 1). The first test uses the multivariate Delta method to estimate the variance of the estimated expected revenue. The second test uses the likelihood-ratio test to make inference. A log-normal distribution is a good candidate to deal with fat tails and is widely used in science and business ([10]). This distribution can arise due to wealth and income distribution ([4]) or the cognitive lock-in of customers to the Staples website ([6]). We checked Staples.com order values data for log-normal distribution using $Q-Q$ plots, and found that it fits (See Appendix).

Method 1: Expected Revenue Confidence Interval Using the Delta Method

To compare expected revenues between treatment and control groups, we will derive a confidence interval for this difference. This confidence interval depends on our estimates for $r$ ($\hat{r}$) and $AOV$ ($\hat{AOV}$), and our uncertainty around them. The estimates $\hat{r}$ and $\hat{AOV}$ are independent. Therefore, their covariance is zero. As a result, we can use the multivariate Delta method222The Delta Method, Theorem 20 of http://www.stat.cmu.edu/~larry/=stat705/Lecture4.pdf, accessed on Feb 21st, 2017. to estimate the variance of $RPV$ from our point and variance estimates of $\hat{r}$ and $\hat{AOV}$. Specifically:

[TABLE]

where all these values are pooled estimates (i.e. the estimates are computed over a sample that consists of combined observations from treatment and control groups).

To estimate $AOV$ and our uncertainty around it from purchase values, which we assumed to have a log-normal distribution, we transform the purchase values by the log function. These transformed values have a normal distribution, so we use its mean $\hat{\mu}$ as a measure for its location, and its standard deviation $\hat{\sigma}_{AOV}$ as a measure for uncertainty around it.

Since $\hat{r}$ is a binomial proportion, its estimated variance is:

[TABLE]

where $\hat{r}$ is the pooled proportions of non-purchases.

As a result, we can write the estimated variance of $RPV$ as follows:

[TABLE]

To make inference, the final step is computing the convergence rate for the asymptotic distribution, which is equal to the harmonic mean of sample sizes:

[TABLE]

As a result, the 95% confidence interval for the expected revenue $RPV$ can be computed as follows:

[TABLE]

Method 2: Likelihood Ratio Test for Expected Revenue

Likelihood ratio test allows comparison between treatment and control groups parametrically. Based on Wilk’s Theorem, the log likelihood ratio has an asymptotic Chi-square distribution with the degree-of-freedom (df) equal to the difference between degrees of freedom of the null and alternative hypotheses. The likelihood ratio test rejects the null hypothesis if the value of this statistic is too small. The Neyman-Pearson lemma states that likelihood ratio test is the most powerful test of significance. Many common test statistics such as Z-test, F-test, Pearson’s chi-square test, and G-test can be phrased as log-likelihood ratio test or its approximation. In order to define the likelihood ratio test for expected revenue, we need to specify a likelihood function under the null and alternative hypotheses. Then we form the likelihood ratio value which is defined as follows:

[TABLE]

where $L$ is the likelihood function. $LR$ is a number between zero and one. The low values of the likelihood ratio mean that the observed result was less likely to occur under the null hypothesis as compared to the alternative.

For our expected revenue test, null hypothesis likelihood function is defined as follows:

[TABLE]

under

[TABLE]

where $OV$ denotes the order value, $LN$ denotes log-normal distribution, and $Bin$ denotes the binomial distribution. The null hypothesis puts constraints over the parameter space of the likelihood function. This constraint requires that the expected revenues under control and treatment groups are equal. Similarly, but excluding the constraint, we can define the likelihood function under the alternative hypothesis as follows:

[TABLE]

In order to run the likelihood ratio test, we need to first find the maximum likelihood estimates of parameters (i.e. $\mu_{c},\sigma_{c}^{2},p_{c},\mu_{t},\sigma_{t}^{2},p_{t}$). To find these parameters for the alternative hypothesis, we can use the closed form solution we specified in the Delta method section because we don’t have any constraints.

As maximizing the likelihood function is equivalent to maximizing the log likelihood function, and derivatives of the log likelihood functions are analytically more tractable than that of the likelihood function, we use the log likelihood function rather than the likelihood function. We define the Lagrangian as follows:

[TABLE]

In order to optimize the Lagrangian, we compute its gradient with respect to the parameter vector (i.e. $(\mu_{c},\sigma_{c}^{2},p_{c},\mu_{t},\sigma_{t}^{2},p_{t},\lambda)$) and equate it to a vector of zeros. As a result, we will have the following system of equations to solve for the parameter vector:

[TABLE]

We did not find a closed form solution for this system of equations, so we use a nonlinear optimization algorithm to numerically find these parameters. Specifying search procedures that gives us this parameter vector completes our specification of this test.

6 Sample Size Estimation

In this section we determine the sample size required in order to obtain the desired statistical power for each of the tests that we discussed above. Note that although we carry two tests sequentially (i.e. Lanchenbruch’s Test followed by either likelihood ratio or Delta method test), the overall sample size we will need for these two tests is going to be smaller than average. This is because, in many cases Lachenbruch’s Two-part Test can establish whether the distribution of active and control groups are the same or not.

The standard process in hypothesis testing consists of collecting data, followed by the execution of the test. In the testing stage, two possible errors can happen: rejecting the null hypothesis when it is true (Type I error), and not rejecting the null hypothesis when it is false (Type II error). In our case, null hypothesis states that the treatment and control groups have the same distribution. As a consequence, a Type I error will make us believe that we have an improvement over our existing system, while in reality we do not. On the other hand, a Type II error makes us believe that the new system is not better than the old system, while in reality it is, so it costs us the opportunity to replace the existing system with a better one.

The probability of committing a Type I error is a function of the shape of the null model distribution. The magnitude of a Type II error is a function of the sample size and the model. Specifying the magnitude of these two types of error allows us to find the sample size required for the test.

Sample size determination using Monte Carlo Method

The Monte Carlo simulation method is applicable for all tests that we discussed above. Steps of this process are as follows:

Run a pilot study to estimate the magnitude of the effect and uncertainty around it. 2. 2.

Specify the desired statistical power level. 3. 3.

Start with an initial guess about the required sample size. 4. 4.

Repeat many times (e.g. 100 or 1000 times) the following:

1. (a)

Sample with replacement from the data collected in the pilot study as many times as the sample size specified in the previous step. 2. (b)

Run the statistical test for each of the samples. 3. (c)

Compute the proportion of times that this test identifies the effect as significant. 4. (d)

Consider this proportion as the power of the statistical test. 5. 5.

Adjust the sample size by comparing the power of the statistical test with the desired power. If you need a higher power, increase the sample size, and run step 2 again.

7 Conclusion, Limitations, and Future Work

In this paper, we addressed the problem of incremental revenue measurement and valid statistical inference. Our approach has three distinct advantages. First, it allows analysts to conclude whether a new system is effective using a smaller sample size. Second, our approach does not require subjective outlier removal. Third, our approach provides a method to determine a confidence interval for $RPV$.

Although our approach improves the current measurement methods, it does not account for the dependence between multiple visits of a given visitor. We address this issue in a forthcoming paper.

Acknowledgements

We thank Karthik Kumar, Stephanie Whang, Ryan Applegate, and Nitin Varma for their helpful discussions.

Appendix- Test of Log-Normality

In order to empirically validate the log-normality of the distribution of revenue per purchase, there are a variety of Normality tests including Jarque-Bera, Kolmogorov-Smirnov, $Q-Q$ plot, and Shapiro-Wilk tests. The latter test (i.e. Shapiro-Wilk test) might not be appropriate for the era of big data333StackOverflow, Is normality test useless , http://stats.stackexchange.com/questions/2492/is-normality-testing-essentially-useless, accessed on Feb 27, 2017.. Other statistical tests such as Kolmogorov-Smirnov and Jarque-Bera are also not appropriate because they raise a red flag with a small deviation from normality, so unless the data has an exactly normal distribution, they return a verdict that the data is not normal444Stackoverflow, Why would all the tests for normality reject the null hypothesis, http://stats.stackexchange.com/questions/16611/why-would-all-the-tests-for-normality-reject-the-null-hypothesis, accessed on Feb 27, 2017. . As a result, we resort to the $Q-Q$ plot approach.

$Q-Q$** **plot test: a quantile-quantile plot compares the cumulative probability density function for two random variables (e.g. empirically-derived distribution vs. known theoretical distribution). It plots the quantiles of two distributions against each other, and in the case where the points lie on the $y=x$ line, it suggests that the distribution of empirical data and theoretical are the same. The advantage of a $Q-Q$ plot over other “goodness of fit” measures is that it is easily recognizable visually. To generate the $Q-Q$ plot we first sort the empirical data into non-decreasing order to yield an inverse probability density function. In addition, we create a vector of indexes from 1 to the number of elements, and calling each index $i$, we create a vector of $\frac{i-0.5}{n}$, where $n$ is the number of observations. We call this the vector of cumulative probability. Next, we create a vector of $z$-scores based on the cumulative probability vector (i.e. the corresponding normally distributed random variable threshold, for which we will use to determine the inverse of the normal distribution). Then we plot the sorted empirical data versus $z$-scores. If there is no systematic deviation from the equality line, then the distribution of empirical data is normal. To test for log-normality, we need to first log transform and standardize the empirical data (i.e. extract mean and divide by standard deviation) and then generate the $Q-Q$ plot555Kratz, M., & Resnick, S. I. (1996). The $Q-Q$-estimator and heavy tails. Stochastic Models, 12(4), 699-724. .