## R exam

Following a long tradition (!) of changing the modus vivendi of each exam in our exploratory statistics with R class, we decided this year to give the students a large collection of exercises prior to the exam and to pick five among them to the exam, the students having to solve two and only two of them. (The exercises are available in French on my webpage.) This worked beyond our expectations in that the overwhelming majority of students went over all the exercises and did really (too) well at the exam! Next year, we will hopefully increase the collection of exercises and also prohibit written notes during the exam (to avoid a possible division of labour among the students).

Incidentally, we found a few (true) gems in the solutions, incl. an harmonic mean resolution of the approximation of the integral

$\int_2^\infty x^4 e^{-x}\,\text{d}x=\Gamma(5,2)$

since some students generated from the distribution with density f proportional to the integrand over [2,∞) [a truncated gamma] and then took the estimator

$\dfrac{1-e^{-2}}{\frac{1}{n}\,\sum_{i=1}^n y_i^{-4}}\approx\dfrac{\int_2^\infty e^{-x}\,\text{d}x}{\mathbb{E}[X^{-4}]}\quad\text{when}\quad X\sim f$

although we expected them to simulate directly from the exponential and average the sample to the fourth power… In this specific situation, the (dreaded) harmonic mean estimator has a finite variance! To wit;

> y=rgamma(shape=5,n=10^5)
> pgamma(2,5,low=FALSE)*gamma(5)
[1] 22.73633
> integrate(f=function(x){x^4*exp(-x)},2,Inf)
22.73633 with absolute error < 0.0017
> pgamma(2,1,low=FALSE)/mean(y[y>2]^{-4})
[1] 22.92461
> z=rgamma(shape=1,n=10^5)
> mean((z>2)*z^4)
[1] 23.92876


So the harmonic means does better than the regular Monte Carlo estimate in this case!

### 2 Responses to “R exam”

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