Suppose that we have $n$ observations : $ x_1, ...,x_n$. ( sort them ). We get $ x_{(1)}, ...,x_{(n)}$

For $\alpha \in [0,\frac{1}{2}) $ we define the $\alpha$-trimmed mean :

$$ T_{\alpha}=\frac{1}{n-2\lceil{\alpha n}\rceil} \sum_{i=\lceil{\alpha n}\rceil +1}^{n-\lceil{\alpha n}\rceil} x_{(i)}$$

and the sample mean is : $$ S= \frac{1}{n} \sum_{i=1}^{n} x_{(i)} $$

$1)$ What happens when a fraction of the observations, say $k$ observations, with $k < \alpha\cdot n$, is huge? Illustrate the effect for $k$ observations tending to infinity on the sample mean and on the $\alpha$-trimmed mean. Explain why the $\alpha$-trimmed mean may be problematic in real life applications compared to the sample mean.

$2)$ Suppose in the setting of part $1)$, that $k \geq 2n \cdot\alpha $. How does this affect the $\alpha$-trimmed mean?

$3)$ Explain why the trimmed mean may be advantageous compared to the sample mean in a real life application, with respect to its robustness.

My idea:

For $1)$ I've considered an example. If we have $n=9$ , $\alpha = 0.3$ ( $k$ has to be $2$ then) with:


Then: $S ≈ 543466$, while $T_{0.3}= 15$. If we now replace $x_{(8)}$ and $x_{(9)}$ by something higher [for example: $400.000$ and $987654321$] (because we want that the $k$ observations tends to infinity ) then we get $S ≈ 109783846$, while $T_{0.3}=15$. So my answer for $1)$ would be that $S$ tending to infinity, while the $k$ observations has no effect on $T_{0.3}$,right? Like you see I have used an example here. Can you help me with the illustration? for the second question of $1)$ I will consider the example above with $\alpha = 0,4$. Then only $x_{(5)}$ would be left. So $T_{0.4} =12 $. So only one observation decides what the trimmed mean is. This is obviously not good, if you want to make a statement of a certain situation. Is this the answer, which is expected?

$2)$ I would say that $T_{\alpha}$ tends to infinity? Because at least one of the observations, which tending to infinity is in $T_{\alpha}$. But my answer is really short? Did I really figure out what the exercise wants to show me?

$3)$ So the trimmed mean is obviously less sensitive to outliers than the sample mean and still illustrates the central tendency. ( see that in $(1)$ ) But to be honest: I don't figured out which advantage "hides" in $(2)$. Can you help me here?

Thank you for your help and correction.


1 Answer 1


Two comments on trimmed means (illustrated in R) that may be helpful.

Fifty percent trimmed mean is median: If the fraction trimmed from each tail is nearly 1/2, the trimmed mean is (nearly) the median. It is not true that only one (or two) observations determine the trimmed mean because the relative positions of the other observations influence which one(s) are not trimmed. Here is an illustration in R.

mean(x); mean(x, trim=.499);  median(x)
[1] 105.6938
[1] 70.74927
[1] 70.74927
[1] 69.90573 70.61226 70.69725 70.80129 70.86220

Robust estimate of mean for contaminated data. In practice, the trimmed mean is especially useful if observed data are mainly from a target distribution, but are subject to 'contamination' by a much different distribution a small percentage of the time. For target distribution $\mathsf{Norm}(100, 15),$ contamination $\mathsf{Norm}(200, 50),$ and contamination is 3%, here is an example:

m = 1000;  av = adj.av = numeric(m)
mu = 100;  sg = 15;  c.mu = 200;  c.sg = 50;  c.rat = .03
b = rbinom(1000, 1, c.rat)
x = (1-b)*rnorm(m, mu, sg) + b*rnorm(m, c.mu, c.sg)
mean(x);  mean(x, trim=.05);  median(x)
[1] 103.9161
[1] 100.9978
[1] 100.4297

enter image description here

  • $\begingroup$ Oh! I totally forget the "relative positions"-point. Thank you for your detailed answer. $\endgroup$ Jun 10, 2018 at 13:18

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