-
Notifications
You must be signed in to change notification settings - Fork 219
/
bayesian_vaccine_effectiveness.Rmd
259 lines (176 loc) · 6.05 KB
/
bayesian_vaccine_effectiveness.Rmd
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
# 贝叶斯分析案例-新冠疫苗有效率的计算 {#bayesian-vaccine-effectiveness}
## 引言
```{r, include=FALSE}
knitr::opts_chunk$set(
echo = TRUE,
warning = FALSE,
message = FALSE,
fig.showtext = TRUE
)
```
```{r eda-vaccine-effectiveness-1, out.width = '70%', fig.align='center', echo = FALSE}
knitr::include_graphics("images/vaccine.png")
```
[纽约时报](https://www.nytimes.com/2020/11/18/health/pfizer-covid-vaccine.html?auth=login-google)报道说,
> 美国制药公司辉瑞(Pfizer)和德国生物科技公司(BioNTech)11月9日率先宣布
,根据在数国临床试验初步结果,其研发的新冠疫苗有效率达到90%以上,星期三,完整结果显示,参加疫苗实验的44000个志愿者中,共有170人确诊感染,其中安慰剂组162人,接种疫苗组仅8人,这证明了辉瑞开发的新冠疫苗有效率高达95%。
```{r eda-vaccine-effectiveness-2, echo=FALSE}
d <- tibble::tribble(
~group, ~volunteers, ~got_covid,
"placebo", 22000L, 162L,
"vaccinated", 22000L, 8L
)
knitr::kable(d)
```
**新冠疫苗是有效的,且有效率高达95%。** 那么,这个95%是怎么计算出来的呢?它的概率是多少以及不确定性是多少呢?
回到这个问题,我们首先需要了解,辉瑞公司是如何定义疫苗有效率的
$$
\text{VE} = 1 - \frac{p_{t}}{p_{c}}
$$
其中$p_t$是**疫苗组**(vaccinated)的感染率,$p_c$是**安慰剂组**(placebo)的感染率。
## 模型
```{r eda-vaccine-effectiveness-3, message=FALSE, warning=FALSE}
library(tidyverse)
library(tidybayes)
library(rstan)
rstan_options(auto_write = TRUE)
options(mc.cores = parallel::detectCores())
```
然后,我们建立如下数学模型:
$$
\begin{align}
y_{c} \sim \textsf{binomial}(n_{c},p_{c}) \\
y_{t} \sim \textsf{binomial}(n_{t},p_{t}) \\
p_{c} \sim \textsf{beta}(1, 1) \\
p_{t} \sim \textsf{beta}(1, 1)
\end{align}
$$
通过模型可以直接计算干预效果$\textsf{effect}$和疫苗有效性$VE$
$$
\begin{align}
\text{effect} = p_{t} - p_{c} \\
\text{VE} = 1 - \frac{p_{t}}{p_{c}}
\end{align}
$$
## 计算
具体Stan代码如下
```{r eda-vaccine-effectiveness-4}
stan_program <- "
data {
int<lower=1> event_c; // num events, control
int<lower=1> event_t; // num events, treatment
int<lower=1> n_c; // num of person trial, control
int<lower=1> n_t; // num of person trial, treatment
}
parameters {
real<lower=0,upper=1> p_c;
real<lower=0,upper=1> p_t;
}
model {
event_c ~ binomial(n_c, p_c);
event_t ~ binomial(n_t, p_t);
p_c ~ beta(1, 1);
p_t ~ beta(1, 1);
}
generated quantities {
real effect = p_t - p_c;
real VE = 1- p_t /p_c;
real log_odds = log(p_t / (1- p_t)) - log(p_c / (1- p_c));
}
"
stan_data <- list(
event_c = 162,
event_t = 8,
n_c = 4.4e4 / 2,
n_t = 4.4e4 / 2
)
mod_vaccine <- stan(model_code = stan_program, data = stan_data)
```
```{r, eval=FALSE, include=FALSE}
# 运行stan代码,导致渲染bookdown报错,不知道为什么,先用这边笨办法凑合吧
# mod_vaccine %>% saveRDS(here::here("stan","mod_vaccine.rds"))
# mod_vaccine <- readRDS(here::here("stan","mod_vaccine.rds"))
```
## 结果
最后,我们后验概率抽样
```{r eda-vaccine-effectiveness-5, eval=FALSE, include=FALSE}
mod_vaccine
```
```{r eda-vaccine-effectiveness-6}
draws <- mod_vaccine %>%
tidybayes::spread_draws(effect, VE, log_odds)
draws %>%
head()
```
### 干预效果
从结果中看到effect中很多负数。事实上,effect中越多的负值,即被感染的可能性越低,说明疫苗干预效果越好
```{r eda-vaccine-effectiveness-7}
mean(draws$effect < 0) %>% round(2)
```
结果告诉我们,疫苗有明显的干预效果。比如,我们假定10000个人接受了疫苗,那么被感染的人数以及相应的可能性,如下图
```{r eda-vaccine-effectiveness-8}
draws %>%
ggplot(aes(x = effect * 1e4)) +
geom_density(fill = "blue", alpha = .2) +
expand_limits(y = 0) +
theme_minimal() +
xlab("效应大小") +
ggtitle("每10000个接种疫苗的人中被感染新冠的数量")
```
```{r eda-vaccine-effectiveness-9, eval=FALSE, include=FALSE}
draws %>%
ggplot(aes(x = log_odds)) +
geom_density(fill = "blue", alpha = .2) +
expand_limits(y = 0) +
theme_minimal() +
xlab("Log odds") +
ggtitle("Log odds of the treatment effect. More negative, less likely to get infected on treatment")
```
### 疫苗有效率
```{r eda-vaccine-effectiveness-10, eval=FALSE, message=FALSE, warning=FALSE, include=FALSE}
draws %>%
ggdist::mean_qi(.width = c(0.95))
draws %>%
ggdist::median_qi(.width = c(0.95))
median(draws$VE)
```
我们再看看疫苗有效率 VE 的结果
```{r eda-vaccine-effectiveness-11}
draws %>%
select(VE) %>%
ggdist::median_qi(.width = c(0.90))
```
通过数据看出,疫苗的有效性为0.95,在90%的可信赖水平, 中位数区间[0.91, 0.97].
```{r eda-vaccine-effectiveness-12, eval=FALSE, include=FALSE}
draws %>%
ggplot(aes(x = VE)) +
geom_density()
draws %>%
ggplot(aes(x = VE)) +
geom_density(fill = "blue", alpha = .2) +
expand_limits(y = 0) +
theme_minimal() +
geom_vline(xintercept = median(draws$VE), size = 0.2)
```
当然,通过图可能理解的更清晰。
```{r eda-vaccine-effectiveness-14}
label_txt <- paste("median =", round(median(draws$VE), 2))
draws %>%
ggplot(aes(x = VE)) +
geom_density(fill = "blue", alpha = .2) +
expand_limits(y = 0) +
theme_minimal() +
geom_vline(xintercept = median(draws$VE), size = 0.2) +
annotate("text", x = 0.958, y = 10, label = label_txt, size = 3) +
xlab("疫苗有效率") +
ggtitle("辉瑞公司定义疫苗有效率为 VE = 1 - Pt/Pc")
```
```{r eda-vaccine-effectiveness-15, echo = F}
# remove the objects
# ls() %>% stringr::str_flatten(collapse = ", ")
#rm(d, draws, label_txt, mod_vaccine, stan_data, stan_program)
rm(d, draws, label_txt, mod_vaccine)
```
```{r eda-vaccine-effectiveness-16, echo = F, message = F, warning = F, results = "hide"}
pacman::p_unload(pacman::p_loaded(), character.only = TRUE)
```