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Using the mills ratio result, let $X \sim N(\mu, \sigma^2)$, then

$E(X| X<\alpha) = \mu - \sigma\frac{\phi(\frac{a- \mu}{\sigma})}{\Phi(\frac{a-\mu}{\sigma})}$

However, when calculating it in R. I don't obtain the correct results as

> mu <- 1
> sigma <- 2 
> a <- 3 
> x <- rnorm(1000000, mu, sigma) 
> x <- x[x < a] 
> mean(x)
[1] 0.4254786
> 
> mu -  sigma * dnorm(a, mu, sigma) / pnorm(a, mu, sigma)
[1] 0.7124

What am I doing wrong?

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1 Answer 1

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Your formula implementation is wrong because, $$\phi\left(\frac{x-\mu}{\sigma}\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\neq f_{X,\mu,\sigma}(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$$ As you can see, we have an extra $\sigma$ in the denominator of $f_{X,\mu,\sigma}(x)$, which yields: $$\phi\left(\frac{x-\mu}{\sigma}\right)=\sigma f_{X,\mu,\sigma}(x)$$ dnorm method gives you $f_{X,\mu,\sigma}(x)$, where you need to multiply it with $\sigma$ to obtain $\phi$. Since your $\sigma=2$, this can be practically done via subtracting the second term again, which is $1-0.7124=0.2876$: $$1-0.2876-0.2876=0.4247$$ which is close to your estimate.

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