16
votes
Accepted
Are there cases where we need to avoid the usage of IQR?
First, there's no reason IQR ought to be close to SD or MAD. While they are all measuring some aspect of dispersion, they are doing it in very different ways. An analogy might be that height, width, ...
11
votes
Accepted
Show that, for any real numbers a and b such that m ≤ a ≤ b or m ≥ a ≥ b, E|Y − a| ≤ E|Y − b| ,where Y be a random variable with finite expectation
Intuition
As explained at Expectation of a function of a random variable from CDF, an integration by parts shows that when a random variable $X$ has a (cumulative) distribution function $F,$ the ...
11
votes
Is mean absolute deviation smaller than standard deviation for $n\ge 3$?
No, in general this is not true.
A simple way to look at this is to simulate. I typically hack together an infinite loop that stops if it finds a counterexample. If it runs for a long time, I start ...
8
votes
Is mean deviation the same as mean absolute difference?
Mean deviation is the same as mean absolute deviation; it is mean deviation from the mean.
$$
MAD=\frac{1}{N}\sum_{i=1}^{N}|x_i-\overline{x}|
$$
Mean absolute difference is for two independent ...
8
votes
Accepted
Random variables $(X,Y)$ with $\text{Var}(X)<\text{Var}(Y)$ and $\mathbb{E}(|X-\mu_X|)>\mathbb{E}(|Y-\mu_Y|)$
Consider a Laplace distribution with scale parameter equal to $1$. One can compute that the MAD equals $1$, and the variance equals $2$.
Now consider a Normal distribution with arbitrary mean, and ...
5
votes
Are there cases where we need to avoid the usage of IQR?
@Peter Flom has given an excellent answer. I want just to gather together a few extra standard comments that deserve some emphasis.
Always plot your data. I often find that box plots emphasizing the ...
5
votes
Random variables $(X,Y)$ with $\text{Var}(X)<\text{Var}(Y)$ and $\mathbb{E}(|X-\mu_X|)>\mathbb{E}(|Y-\mu_Y|)$
Consider a discrete random variable $Y$ that, for some constants $M\in\mathbb{R}$ and $\epsilon\in [0, 1]$, takes the following values:
$$
Y = \left\{\begin{array} ~-M+\mu_Y & \text{w.p.}~\epsilon/...
5
votes
Why don't dispersions like median deviation and mode deviation exists on the lines of mean deviation?
Yes we can, there is, for example, median absolute deviation (MAD):
$$ \operatorname{MAD}(X) = \operatorname{median}\left(\ \left| X_{i} - \operatorname{median} (X) \right|\ \right)$$
It has even ...
5
votes
Accepted
an upper bound of mean absolute difference?
Theorem 3.3 from p. 86 of "Cerone, Pietro, and Sever S. Dragomir. "A survey on bounds for the Gini Mean Difference." Advances in Inequalities from Probability Theory and Statistics (...
5
votes
Why not use modulus for variance?
Let $\mu=\operatorname{E}(X).$
The main reason for using $\sqrt{\operatorname{var}(X)} = \sqrt{\operatorname{E}((X-\mu)^2)}$ as a measure of dispersion, rather that using the mean absolute deviation $...
5
votes
Average deviation from mean of Irwin-Hall distribution
I am not quite clear what you are asking for.
For $n=1$ you get $\mathbb E[|X_1|]=\frac14=0.25\times \sqrt{1}$.
For $n=2$ you get $\mathbb E[|X_2|]=\frac13\approx 0.2357\times \sqrt{2}$.
The ...
4
votes
Why do we use squared deviations to compute the SD, given that it amplifies the effect of outliers?
I have read in many places that one of the reasons we take the square
for calculating the SD is because we want to give more weight to
outliers.
This is not correct. The use of the square in ...
4
votes
Why is there no improvement when training Xgboost with pseudo-Huber loss?
There is no definite answer at this but I would note one major and one minor point:
The major point is that: A XGBoost booster starts with a base_score. That is ...
4
votes
How to calculate the running mean absolute deviation
There is no (useful) exact recursion, but there is an approximate one
Since your comment specified that you prefer to use the MAD around the mean, I will proceed for that case. Given an observed ...
4
votes
Average deviation of a set of 10 random integers from −5 to +5
Your title and text seems to ask about the expectation of the mean deviation of a sample size of $N=10$ random integers from -5 to 5. A secondary question (for which I don't supply an answer) seems ...
3
votes
Accepted
Upper bound on absolute difference with squares
I would have thought you could say:
$E\big|X^2-(E[X])^2\big| \le E\big|X^2+(E[X])^2\big| = E\big[X^2\big]+(E[X])^2$
As an example to show it can come close:
suppose $X=1000$ with probability $\frac1{...
3
votes
Why does Mean deviation = Standard deviation = Range/2?
They aren't equal in general, as Gomunkul points out; in the two observation case they are; that's the situation your book is referring to.
Since you already have a proof, my best guess is that you'...
3
votes
Why does Mean deviation = Standard deviation = Range/2?
Without going to the R(Range)/2 part, why do you think MD = SD?
That is NOT generally true (although data can be constructed such that MD = SD).
For example: x = [-10, 0, 10]
SD(x) = sqrt((10^2 + 0 + ...
2
votes
Variance of Binomial total deviation
Let $\mu_X=E(X)=np$. Call the mean deviation, $E|X-\mu|$,
"$m_X$"
$\text{Var}(|X-\mu|) = E(|X-\mu|^2)- (E|X-\mu|)^2 = \text{Var}(X) - m_X^2$.
$\text{Var}(X)$ for a binomial$(n,p)$ is easily obtained ...
2
votes
Is mean absolute deviation smaller than standard deviation for $n\ge 3$?
Here is a more mathematical approach. Firstly, it's probably true that by a change of variables, one can assume that the mean is zero. Certainly from the point of view of finding a counter example, ...
2
votes
Accepted
What are the drawbacks of aggregating regression results if error decreases?
This looks like a hierarchical forecasting problem. One issue is that you need to be careful when evaluating MAE for state level and national level. MAE depends on the absolute value of the target ...
2
votes
Alternative error measure--maximize data to have residuals less than given threshold
This is approximately what support vector regression does.
Minimize $${\displaystyle {\frac {1}{2}}\|\beta\|^{2}}$$
subject to $${\displaystyle |y_{i}-x_i\beta -\beta_0|\leq \varepsilon }.$$
That is, ...
2
votes
Absolute deviation from the mean using logarithms
If your $X$ data are log-normally distributed and you want to use a measure like $\left| \ln(x_i) - C\right |$ to describe the spread of the data, where $C$ is some measure of central tendency, then ...
2
votes
Is it incorrect to take the standard deviation of absolute values
You can calculate the standard deviation of whatever you want, the normal distribution has nothing to do with it. Standard deviation is the square root of the mean squared error. If your errors were ...
1
vote
MAD/Mean Ratio - advantages/disadvantages of using average or sum
If you are a little more careful and explicit in your formulas, it might well get a lot clearer.
Specifically, one typically calculates means or sums over forecasts and actuals for the same periods (...
1
vote
Mean absolute deviation (MAD) analogy to 68-95-99 rule
Assuming Normally distributed data (since the $68-95-99$ rule comes from the Normal distribution) I estimate that about $82\%$ of samples lie within $\pm 2\text{MAD}$ of the median and around $96\%$ ...
1
vote
Accepted
measure for deviation/error that is guarantueed to be smaller than the mean
So, you know that the true values of this unknown value must be greater than zero? This seems like the perfect time to incorporate a prior.
Let's call the random variable $X$. Since we know $X$ cannot ...
1
vote
Accepted
Ho do I determine the probability of price reversals using linear regression?
Linear regression is a model that can predict values in unrestricted domain, from $-\infty$ to $\infty$. If you want a regression model to predict probabilities, then the to-go model is logistic ...
1
vote
Why not use modulus for variance?
There are already several good answers here, including in the comments. However as the OP requested a "simpler" justification, here I will expand on my comment.
To me this is a very natural ...
1
vote
Estimating Expected Order Statistics
I would like find an estimator $\hat{μ}_{k:n}$ which takes a profile of $m$ iid
samples and minimizes the mean absolute error $\mathbb{E}[|\hat{μ}_{k:n}−μ_{k:n}|]$ in the
worst case over all $F$
...
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