Example - yearly wage of 474 bank employees
Data provided by Professor Chris Skeels in Econometrics 3 ECOM90013
H0:βeduc=0
H1:βeduc≠0
## LM Testlm0 <- lm(LOGSAL ~ GENDER + MINORITY + JOBCAT, data = wages)e0 <- residuals(lm0)lm1 <- lm(e0 ~ EDUC + GENDER + MINORITY + JOBCAT, data = wages)e1 <- summary(lm1)e1rsq <- e1$r.squaredtest1 <- nrow(wages) * e1rsq
```{r, echo = FALSE, result = 'asis'}cat( "Under the null hypothesis with degree of freedom equal to 1,", " the test statistic is ",round(test1,4), " and critical value is ", round(qchisq(0.95,1),4))```
Under the null hypothesis with degree of freedom equal to 1 , the test statistic is 125.7683 and the critical value is 3.8415.
H0:βeduc=0
H1:βeduc≠0
reject_h0 <- test1 > round(qchisq(0.95, 1), 4)
Since the test statistic for LM1 is `r if(reject_h0) "greater" else "smaller" `
greater than the critical, therefore we `r if(reject_h0) "" else " cannot" `
reject the null hypothesis and conclude that βeduc is `r if(reject_h0) "" else " not"`
significant at 5% level.
Since the test statistic for LM1 is greater than the critical, therefore we reject the null hypothesis and conclude that βeduc is significant at 5% level.
H0:βminority=
βjobcat=0
H1:βminority≠0
or βjobcat≠0
lmrest <- lm(formula = LOGSAL ~ EDUC + GENDER, data = wages)e2 <- summary(lmrest)$residualslme2 <- lm(e2 ~ EDUC + GENDER + MINORITY + JOBCAT, data = wages)e2.sqr <- summary(lme2)$r.squaredtest2 <- nrow(wages) * e2.sqrprint("Under the null hypothesis with degree of freedom equal to 2")
## [1] "Under the null hypothesis with degree of freedom equal to 2"
print(paste0("the test statistic is ", round(test2, 4)))
## [1] "the test statistic is 208.745"
print(paste0("The critical value is ", round(qchisq(0.95, 2), 4)))
## [1] "The critical value is 5.9915"
H0:βminority=
βjobcat=0
H1:βminority≠0
or βjobcat≠0
reject_h0.2 <- test2 > round(qchisq(0.95, 2), 4)
Since the test statistic for LM1 is `r if(reject_h0.2) "greater" else "smaller"`
greater than the critical, therefore we `r if(reject_h0.2) "" else " cannot" `
reject the null hypothesis and conclude that `r if(reject_h0.2) "at least one of" else "none of" `
at least one of βminority and βjobcat is significant at 5% level.
Since the test statistic for LM1 is greater than the critical, therefore we reject the null hypothesis and conclude that at least one of βminority and βjobcat is significant at 5% level.
Bayes' Rule: p(θ|Y)∝L(θ|Y)p(θ)
The posterior distribution is proportion to the kernel of posterior distribution times the distribution of the prior distribution.
Bayes' Rule: p(θ|Y)∝L(θ|Y)p(θ)
The posterior distribution is proportion to the kernel of posterior distribution times the distribution of the prior distribution.
We have a time series for Australian real GDP from the Australian Real-Time Macroeconomic Database containing T=230 observations on the quarterly data from quarter 3 of 1959 to the last quarter of 2016.
Data provided by Tomasz Wozniak in Macroeconometrics ECOM90007
Question: "Set the parameters of the natural-conjugate prior distribution and motivate the values that you choose."
Random Walk with drift process: logGDPt=μ0+αlogGDPt−1+ut
α=1
ut∼N(0,σ2)
P(σ2)∼IG2(s,ν)
Priors: μ0, α, σ2, s, ν
Question: "Set the parameters of the natural-conjugate prior distribution and motivate the values that you choose."
Random Walk with drift process: logGDPt=μ0+αlogGDPt−1+ut
α=1
ut∼N(0,σ2)
P(σ2)∼IG2(s,ν)
Priors: μ0, α, σ2, s, ν
The sample mean of μ0 with 5000 draws is 0.0148564 and the variance is 0.011913.
The sample mean of α with 5000 draws is 0.999454 and the variance is 0.000082.
The sample mean of σ2 with 5000 draws is 0.017256 and the variance is 0.0000026.
The sample mean of μ0 with 5000 draws is 0.0024582 and the variance is 0.001686.
The sample mean of α with 5000 draws is 1.00048 and the variance is 0.000012.
The sample mean of σ2 with 5000 draws is 0.017258 and the variance is 0.0000026.
The sample mean of μ0 with 5000 draws is 0.0114882 and the variance is 0.011913.
The sample mean of α with 5000 draws is 0.999733 and the variance is 0.000082.
The sample mean of σ2 with 5000 draws is 0.017257 and the variance is 0.0000026.
The sample mean of μ0 with 5000 draws is `r round(mean(blogau$V1),8)`
and the variance is `r round(var(blogau$V1),6)`
.
The sample mean of α with 5000 draws is `r round(mean(blogau$V2),6)`
and the variance is `r round(var(blogau$V2),8)`
.
The sample mean of σ2 with 5000 draws is`r round(mean(blogau$sigmasq),6)`
and the variance is `r round(var(blogau$sigmasq),8)`
.
shh, witchcraft here. Why do I need this chunk to advance to next slide?@yihui, please send help!
pdf(file="mu0plot.pdf", height=12, width=9)ggplot(data=blogau, aes(x=V1)) + geom_histogram(binwidth=0.01, colour="black", fill="white")+ ggtitle("Distribution of mu0")+ xlab("mu0") dev.off()
pdf(file="mu0plot.pdf", height=12, width=9)ggplot(data=blogau, aes(x=V1)) + geom_histogram(binwidth=0.01, colour="black", fill="white")+ ggtitle("Distribution of mu0")+ xlab("mu0") dev.off()
```{r,echo=FALSE,fig.height=12,fig.width=9,dev="pdf"}ggplot(data=blogau, aes(x=V1)) + geom_histogram(binwidth=0.01, colour="black", fill="white")+ ggtitle("Distribution of mu0")+ xlab("mu0") ```
Alison Hill, June 2019, R-Ladies xaringan theme:
Professor Chris Skeels, S1 2020,Econometrics ECOM90013, University of Melbourne
Guidotti, E., Ardia, D., (2020), "COVID-19 Data Hub", Journal of Open Source Software 5(51):2376, doi:10.21105/joss.02376.
Tomasz Wozniak, S1 2020, Macroeconometrics ECOM90007, University of Melbourne
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https://www.linkedin.com/in/danyang-dai-7529b4152/
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