The paper
presents Pareto smoothed importance sampling, but also Pareto-\(\hat{k}\) diagnostic that can be used when
estimating any expectation based on finite sample. This vignette
illustrates the use of these diagnostics. The individual diagnostic
functions are pareto_khat(), pareto_min_ss(),
pareto_convergence_rate() and
pareto_khat_threshold(). The function
pareto_diags() will return all of these.
Additionally, the pareto_smooth() function can be used
to transform draws by smoothing the tail(s). ## Example
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options(pillar.neg = FALSE, pillar.subtle=FALSE, pillar.sigfig=2)Generate xn a simulated MCMC sample with 4 chains each
with 1000 iterations using AR process with marginal normal(0,1)
N <- 1000
phi <- 0.3
set.seed(6534)
dr <- array(data=replicate(4,as.numeric(arima.sim(n = N,
list(ar = c(phi)),
sd = sqrt((1-phi^2))))),
dim=c(N,4,1)) %>%
as_draws_df() %>%
set_variables('xn')Transform xn via cdf-inverse-cdf so that we have
variables that have marginally distributions \(t_3\), \(t_{2.5}\), \(t_2\), \(t_{1.5}\), and \(t_1\). These all have thick tails. In
addition \(t_2\), \(t_{1.5}\), and \(t_1\) have infinite variance, and \(t_1\) (aka Cauchy) has infinite mean.
We examine the draws with the default
summarise_draws().
drt %>%
summarise_draws()## # A tibble: 6 × 10
## variable mean median sd mad q5 q95 rhat ess_bulk ess_tail
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.010 0.00094 0.99 0.99 -1.6 1.6 1.0 2284. 3189.
## 2 xt3 0.025 0.0010 1.6 1.1 -2.3 2.3 1.0 2284. 3189.
## 3 xt2_5 0.031 0.0010 2.0 1.1 -2.5 2.5 1.0 2284. 3189.
## 4 xt2 0.046 0.0011 2.9 1.2 -2.8 2.9 1.0 2284. 3189.
## 5 xt1_5 0.092 0.0011 7.6 1.3 -3.5 3.6 1.0 2284. 3189.
## 6 xt1 0.33 0.0012 93. 1.5 -5.8 6.1 1.0 2284. 3189.All the usual convergence diagnostics \(\widehat{R}\), Bulk-ESS, and Tail-ESS look good, which is fine as they have been designed to work also with infinite variance (Vehtari et al., 2020).
If these variables would present variables of interest for which we would like to estimate means, we would be also interested in Monte Carlo standard error (MCSE, see case study How many iterations to run and how many digits to report).
drt %>%
summarise_draws(mean, sd, mcse_mean, ess_bulk, ess_basic)## # A tibble: 6 × 6
## variable mean sd mcse_mean ess_bulk ess_basic
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.010 0.99 0.021 2284. 2280.
## 2 xt3 0.025 1.6 0.033 2284. 2452.
## 3 xt2_5 0.031 2.0 0.039 2284. 2584.
## 4 xt2 0.046 2.9 0.054 2284. 2903.
## 5 xt1_5 0.092 7.6 0.13 2284. 3553.
## 6 xt1 0.33 93. 1.5 2284. 3976.Here MCSE for mean is based on standard deviation and Basic-ESS, but these assume finite variance. We did sample also from distributions with infinite variance, but given a finite sample size, the empirical variance estimates are always finite, and thus we get overoptimistic MCSE.
To diagnose whether our variables of interest may have infinite variance and even infinite mean, we can use Pareto-\(\hat{k}\) diagnostic.
drt %>%
summarise_draws(mean, sd, mcse_mean, ess_basic, pareto_khat)## # A tibble: 6 × 6
## variable mean sd mcse_mean ess_basic pareto_khat
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.010 0.99 0.021 2280. -0.072
## 2 xt3 0.025 1.6 0.033 2452. 0.33
## 3 xt2_5 0.031 2.0 0.039 2584. 0.41
## 4 xt2 0.046 2.9 0.054 2903. 0.52
## 5 xt1_5 0.092 7.6 0.13 3553. 0.69
## 6 xt1 0.33 93. 1.5 3976. 1.0\(\hat{k} \leq 0\) indicates that
all moments exist, and the inverse of positive \(\hat{k}\) tells estimate for the number of
finite (fractional) moments. Thus, \(\hat{k}\geq 1/2\) indicates infinite
variance, and \(\hat{k}\geq 1\)
indicates infinite mean. Sometimes very thick distribution tails may
affect also sampling, but assuming sampling did go well, and we would be
interested only in quantiles, infinite variance and mean are not a
problem. But if we are interested in mean, then we need to care about
the number of (fractional) moments. Here we see \(\hat{k} \geq 1/2\) for \(t_2\), \(t_{1.5}\), and \(t_{1}\), and we should not trust their
mcse_mean values. Without trustworthy MCSE estimate we
don’t have good estimate of how accurate the mean estimate is.
Furthermore, as \(\hat{k} \geq 1\) for
\(t_{1}\), the mean is not finite and
the mean estimate is not valid.
If we really do need those mean estimates, we can improve trustworthiness by Pareto smoothing, which replaces extreme tail draws with expected ordered statistics of Pareto distribution fitted to the tails of the distribution. Pareto smoothed mean estimate (computed using Pareto smoothed draws) has finite variance with a cost of some bias which we know when it is negligible. As a thumb rule when \(\hat{k}<0.7\), the bias is negligible.
We do Pareto smoothing for all the variables.
drts <- drt %>%
mutate_variables(xt3_s=pareto_smooth(xt3),
xt2_5_s=pareto_smooth(xt2_5),
xt2_s=pareto_smooth(xt2),
xt1_5_s=pareto_smooth(xt1_5),
xt1_s=pareto_smooth(xt1)) %>%
subset_draws(variable="_s", regex=TRUE)## Pareto k-hat = 0.32.## Pareto k-hat = 0.4.## Pareto k-hat = 0.51.## Pareto k-hat = 0.68.## Pareto k-hat = 1.02. Mean does not exist, making empirical mean estimate of the draws not applicable.Now the mcse_mean values are more trustworthy when \(\hat{k} < 0.7\). When \(\hat{k}>0.7\) both bias and variance
grow so fast that Pareto smoothing rarely helps (see more details in the
paper).
drts %>%
summarise_draws(mean, mcse_mean, ess_basic, pareto_khat)## # A tibble: 5 × 5
## variable mean mcse_mean ess_basic pareto_khat
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 xt3_s 0.026 0.033 2438. 0.33
## 2 xt2_5_s 0.033 0.038 2536. 0.40
## 3 xt2_s 0.052 0.051 2763. 0.50
## 4 xt1_5_s 0.12 0.10 3293. 0.67
## 5 xt1_s 0.97 0.80 3903. 0.98The bias and variance depend on the sample size, and we can use
additional diagnostic min_ss which tells the minimum sample
size needed so that mcse_mean can be trusted.
drt %>%
summarise_draws(mean, mcse_mean, ess_basic,
pareto_khat, min_ss=pareto_min_ss)## # A tibble: 6 × 6
## variable mean mcse_mean ess_basic pareto_khat min_ss
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.010 0.021 2280. -0.072 10
## 2 xt3 0.025 0.033 2452. 0.33 31.
## 3 xt2_5 0.031 0.039 2584. 0.41 48.
## 4 xt2 0.046 0.054 2903. 0.52 116.
## 5 xt1_5 0.092 0.13 3553. 0.69 1735.
## 6 xt1 0.33 1.5 3976. 1.0 InfHere required min_ss is smaller than
ess_basic for all except \(t_1\), for which there is no hope.
Given finite variance, the central limit theorem states that to halve MCSE we need four times bigger sample size. With Pareto smoothing, we can go further, but the convergence rate decreases when \(\hat{k}\) increases.
drt %>%
summarise_draws(mean, mcse_mean, ess_basic,
pareto_khat, min_ss=pareto_min_ss,
conv_rate=pareto_convergence_rate)## # A tibble: 6 × 7
## variable mean mcse_mean ess_basic pareto_khat min_ss conv_rate
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.010 0.021 2280. -0.072 10 1
## 2 xt3 0.025 0.033 2452. 0.33 31. 0.98
## 3 xt2_5 0.031 0.039 2584. 0.41 48. 0.95
## 4 xt2 0.046 0.054 2903. 0.52 116. 0.86
## 5 xt1_5 0.092 0.13 3553. 0.69 1735. 0.60
## 6 xt1 0.33 1.5 3976. 1.0 Inf 0We see that with \(t_2\), \(t_{1.5}\), and \(t_1\) we need \(4^{1/0.86}\approx 5\), \(4^{1/0.60}\approx 10\), and \(4^{1/0}\approx \infty\) times bigger sample sizes to halve MCSE for mean.
The final Pareto diagnostic, \(\hat{k}\)-threshold, is useful for smaller sample sizes. Here we select only 100 iterations per chain to get total of 400 draws.
drt %>%
subset_draws(iteration=1:100) %>%
summarise_draws(mean, mcse_mean, ess_basic,
pareto_khat, min_ss=pareto_min_ss,
khat_thres=pareto_khat_threshold,
conv_rate=pareto_convergence_rate)## # A tibble: 6 × 8
## variable mean mcse_mean ess_basic pareto_khat min_ss khat_thres conv_rate
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xn 0.038 0.066 244. -0.012 1 e 1 0.62 1
## 2 xt3 0.054 0.11 237. 0.32 3.0e 1 0.62 0.96
## 3 xt2_5 0.057 0.13 237. 0.38 4.2e 1 0.62 0.93
## 4 xt2 0.063 0.17 243. 0.48 8.3e 1 0.62 0.86
## 5 xt1_5 0.061 0.31 271. 0.64 5.6e 2 0.62 0.66
## 6 xt1 -0.26 1.6 344. 0.95 2.2e18 0.62 0.11With only 400 draws, we can trust the Pareto smoothed result only
when \(\hat{k}<0.62\). For \(t_{1.5}\) \(\hat{k}\approx 0.64\), and
min_ss reveals we would probably need more than 560 draws
to be on the safe side.
We can get all these diagnostics with pareto_diags(),
and it’s easy to use it also for derived quantities.
drt %>%
mutate_variables(xt2_5_sq=xt2_5^2) %>%
subset_draws(variable="xt2_5_sq") %>%
summarise_draws(mean, mcse_mean,
pareto_diags)## # A tibble: 1 × 7
## variable mean mcse_mean khat min_ss khat_threshold convergence_rate
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 xt2_5_sq 3.9 0.56 0.67 1124. 0.72 0.63All these diagnostics are presented in Section 3 and summarized in Table 1 in PSIS paper (Vehtari et al., 2024).
If you don’t need to estimate means of thick tailed distributions, and there are no sampling issues due to thick tails, then you don’t need to check existence of finite variance, and thus there is no need to check Pareto-\(\hat{k}\) for all the parameters and derived quantities.
It is possible that the distribution has finite variance, but pre-asymptotically given a finite sample size the behavior can be similar to infinite variance. Thus the diagnostic is useful even in cases where theory guarantees finite variance.
Vehtari, A., Simpson, D., Gelman, A., Yao, Y., & Gabry, J. (2024). Pareto smoothed importance sampling. Journal of Machine Learning Research, 25(72):1-58.
Vehtari A., Gelman A., Simpson D., Carpenter B., & Bürkner P. C. (2020). Rank-normalization, folding, and localization: An improved Rhat for assessing convergence of MCMC. Bayesian Analysis, 16(2):667-718.