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# 统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

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## 统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

If the regression model $\ln (Y)=\beta_0+\beta_1 X+\varepsilon$ is true, then (as discussed above) the conditional mean and variance of the untransformed $Y$ variable are given as follows:
$$\mathrm{E}(Y \mid X=x)=\exp \left(\beta_0+\beta_1 x\right) \times \exp \left(\sigma^2 / 2\right)$$
and
$$\operatorname{Var}(Y \mid X=x)=\left{\exp \left(\sigma^2\right)-1\right} \times \exp \left(\sigma^2\right) \times \exp \left(2 \beta_0+2 \beta_1 x\right)$$

Notice that when $\sigma^2=0$, then $\left{\exp \left(\sigma^2\right)-1\right}=0$, so that $\operatorname{Var}(Y \mid X=x)=0$, as expected. Notice also that if you assume the lognormal model, you also assume that the variance of $Y \mid X=x$ is nonconstant; i.e., heteroscedastic: The formula for $\operatorname{Var}(Y \mid X=x)$ in the lognormal case shows that, if $\beta_1>0$, then the variance increases for larger $X=x$.

Figures 5.9 and 5.10 demonstrate the heteroscedasticity of the lognormal regression model, as well as the homoscedasticity of the model for the log-transformed data.
R code for Figures 5.9 and 5.10
$\mathrm{n}=100 ;$ beta $=-2.00 ;$ betal $=0.05 ;$ sigma $=0.30$
set. seed $(12345) ; X=\operatorname{rnorm}(n, 70,10)$
$\ln Y=\operatorname{beta0}+\operatorname{beta} 1 * X+\operatorname{rnorm}(n, 0, \operatorname{sigma}) ; Y=\exp (\ln Y)$
$\operatorname{par}(m f r o w=c(1,2)) ;$ plot $(X, Y) ; \quad \operatorname{abline}(\mathrm{V}=\mathrm{c}(60,80)$, col=”gray”)
plot $(\mathrm{X}, \ln Y)$; abline $(\mathrm{v}=\mathrm{c}(60,80), \mathrm{col}=$ “gray” $)$
$y \cdot s e q=\operatorname{seq}(0.01,30, .01)$
$d y 1=d \ln \operatorname{drm}(\mathrm{y} \cdot$ seq, beta $0+$ beta $1 * 60$, sigma)
$d y 2=d \operatorname{lnorm}(y \cdot$ seq, beta $0+$ beta $1 \star 80$, sigma)
par (mfrow $=\mathrm{c}(1,2))$
plot (y.seq, dy1, type=”1″, $x \lim =c(0,20)$, yaxs=”in, ylim $=c(0, .6)$,
$y l a b=$ “lognormal density”, $x l a b=$ “Untransformed $y^{\prime \prime}$ )
points(y.seq, dy2, type=”l”, lty $=2$ )
legend (“topright”, $c(” X=60 “, ” X=80 “)$, lty $=c(1,2), c e x=0.8)$
$l y \cdot s e q=\log (y \cdot s e q)$
dly $1=\operatorname{dnorm}(1 y \cdot s e q$, beta0+beta $1 * 60$, sigma)
dly $2=\operatorname{dnorm}(l y \cdot$ seq, beta0+beta $1 \star 80$, sigma)
plot (ly.seq, dlyl, type=”l”, $x l i m=c(-.2,4)$, yaxs=”i”, ylim $=c(0,1.6)$,
$y l a b=$ “normal density”, $x l a b=$ “Log Transformed $y “)$
points(ly.seq, dly2, type=”l”, lty=2)
legend (“topright”, $c(” \mathrm{X}=60 “, ” \mathrm{X}=80 “)$, lty=c(1,2), cex=0.8)

## 统计代写|回归分析代写Regression Analysis代考|An Example Where the Inverse Transformation $1 / Y$ Is Needed

With ratio data, the units of measurement are ( $a$ per $b$ ), and the inverse transformation often makes sense simply because the measurements become ( $b$ per $a)$, which is just as easy to interpret. For example, a car that gets 30 miles per gallon of gasoline equivalently can be stated to take $(1 / 30)$ gallons per mile. You could use either measure in a statistical analysis, without question from any critical reviewer-miles per gallon and gallons per mile convey the same information. Which form to use? Simply choose the form that least violates the model assumptions.

The following code replicates the analyses shown in Figure 5.6, for these data, but using the $W=1 / Y$ transformation, which he called “speed”, because higher values indicate a speedier computer.
R code for Figure 5.11
URA-Datasets/master/compspeed.txt “)
attach (comp)
reg.orig $=1 \mathrm{~m}(t i m e \sim \mathrm{GB}) ;$ summary (reg.orig)
par(mfrow=c(2,2)); plot(GB, time); add.loess (GB, time)
qqnorm(reg.orig\$residuals); qqline(reg.orig\$residuals)
speed $=1 /$ time; reg.trans $=1 \mathrm{~m}$ (speed $\sim$ GB)
summary (reg.trans)
plot (GB, speed); add.loess (GB, speed)
qqnorm(reg.trans\$residuals); qqline(reg.trans\$residuals)
comp $=$ read.table (“https://raw.githubusercontent. com/andrea2 $719 /$
URA-Datasets/master/compspeed.txt “)
attach (comp)
reg.orig $=1 \mathrm{~m}($ time $\sim$ GB) ; $\operatorname{summary}($ reg.orig)
par (mfrow=c(2,2)); plot (GB, time); add.loess (GB, time)
qqnorm(reg.orig\$residuals); qqline (reg.orig\$residuals)
speed $=1 /$ time; reg.trans $=1 \mathrm{~m}($ speed $\sim \mathrm{GB})$
summary (reg.trans)
plot (GB, speed); add.loess (GB, speed)
qqnorm (reg.trans\$residuals) ; qqline (reg.trans\$residuals)

## 统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

$$\mathrm{E}(Y \mid X=x)=\exp \left(\beta_0+\beta_1 x\right) \times \exp \left(\sigma^2 / 2\right)$$

$$\operatorname{Var}(Y \mid X=x)=\left{\exp \left(\sigma^2\right)-1\right} \times \exp \left(\sigma^2\right) \times \exp \left(2 \beta_0+2 \beta_1 x\right)$$

$\mathrm{n}=100 ;$ beta $=-2.00 ;$ betal $=0.05 ;$ sigma $=0.30$

$\ln Y=\operatorname{beta0}+\operatorname{beta} 1 * X+\operatorname{rnorm}(n, 0, \operatorname{sigma}) ; Y=\exp (\ln Y)$
$\operatorname{par}(m f r o w=c(1,2)) ;$ plot $(X, Y) ; \quad \operatorname{abline}(\mathrm{V}=\mathrm{c}(60,80)$, col=”gray”)

$y \cdot s e q=\operatorname{seq}(0.01,30, .01)$
$d y 1=d \ln \operatorname{drm}(\mathrm{y} \cdot$ seq， β $0+$ β $1 * 60$, sigma)
$d y 2=d \operatorname{lnorm}(y \cdot$ seq， β $0+$ β $1 \star 80$, sigma)
Par (mfrow $=\mathrm{c}(1,2))$)
Plot (y.seq, dy1, type=”1″， $x \lim =c(0,20)$, yaxs=”in, ylim $=c(0, .6)$，
$y l a b=$ “对数正态密度”，$x l a b=$ “未转换$y^{\prime \prime}$)

$l y \cdot s e q=\log (y \cdot s e q)$
Dly $1=\operatorname{dnorm}(1 y \cdot s e q$， β a0+ β $1 * 60$, sigma)
Dly $2=\operatorname{dnorm}(l y \cdot$ seq， β a0+ β $1 \star 80$, sigma)

$y l a b=$ “正常密度”，$x l a b=$ “对数变换$y “)$

## 统计代写|回归分析代写Regression Analysis代考|An Example Where the Inverse Transformation $1 / Y$ Is Needed

“URA-Datasets/master/compspeed.txt”)

reg。origin $=1 \mathrm{~m}(t i m e \sim \mathrm{GB}) ;$摘要(reg.origin)
par(mfrow=c(2,2));plot(GB，时间);添加黄土(GB，时间)
qqnorm(reg. origin ＄residuals);qqline(reg. origin ＄残差)

plot (GB，速度);添加黄土(GB，速度)
qqnorm(reg.trans＄residuals);腾讯网(reg.trans＄残差)
Comp $=$阅读。表(“https://raw.githubusercontent。com andrea2 $719 /$
“URA-Datasets/master/compspeed.txt”)

reg。origin $=1 \mathrm{~m}($ time $\sim$ GB);$\operatorname{summary}($ reg.org)
Par (mfrow=c(2,2));plot (GB，时间);添加黄土(GB，时间)
qqnorm(reg. origin ＄residuals);Qqline (reg. origin ＄残差)

plot (GB，速度);添加黄土(GB，速度)
Qqnorm (reg.trans＄残差);腾讯线(reg.trans＄残差)

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