11
votes
Percentage changes versus absolute changes when comparing rankings
This question reminds me of this xkcd strip. All humor aside, comparing percentages which do not share the same basis (i.e. denominator) can be very misleading (as xkcd points out).
In addition, ...
- 5,294
11
votes
Why square the difference instead of taking the absolute value in standard deviation?
Why square the difference instead of taking the absolute value in standard deviation?
We square the difference of the x's from the mean because the Euclidean distance proportional to the square root ...
- 574
10
votes
What is the expectation of the absolute value of the Skellam distribution?
It's possible to write the expectation in terms of easy-to-compute special functions.
Let $z$ follow a Skellam distribution with rates $\lambda_1$ and $\lambda_2$, and $k = |z|$. The pmf for $k$ is:
$...
- 1,136
10
votes
Accepted
Expected value of the absolute standardized t distribution
Per whuber's answer to Standardized Student's-t distribution, the density of the standardized $t$ distribution on $\nu$ degrees of freedom is
$$f(x) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\...
- 131k
9
votes
Accepted
Expected absolute deviation greater than standard Laplace
Consider a ReflectedGamma($b$, $c$) distribution (also known as a double Gamma distribution) with pdf $f(x)$:
$$f(x) = \frac{ |x|^{c-1} \space e^{-|x|/b}}{ 2 \space \Gamma (c) \space \space b^c} \quad ...
- 8,001
8
votes
Accepted
6
votes
Accepted
Expectation of sum of absolute values for correlated normal random variables
Expectation operator distributes over the sum, so correlation amongst $y_i$ is not important in the calculation:
$$\mathbb E\left[\sum_{i=1}^N|y_i|\right]=N\mathbb E[|y_1|]$$
$y_1\sim\mathcal N(0, \...
- 58.2k
6
votes
Accepted
Finding the value of $k$ for an Uniform Distribution defined on $(-k,k)$
Your result is correct, but the method has a small flaw.
Your density should be
$$
f(x) = \frac{1}{2k} \cdot I_{(-k,k)} (x)
$$
This didn't change anything, since both your integration areas lie in $(-...
- 86
4
votes
Distribution of errors
You say you have sufficient proof that $Z$ is normal $\mathcal{N}(\varepsilon,\sigma^2)$.
Evaluate the goodness of fit
You can use the log-likelihood of the data $\mathcal{D}= \{z_1, z_2, \dots, z_n\}$...
- 1,419
4
votes
Representation of the expectation of absolute value of the difference $Y-X$
As @whuber has pointed out, the expressions in your post are true for non-negative random variables, not in general.
For a random variable $X$ with distribution function $G$, one has in general
$$E(X)=...
- 11.6k
3
votes
Accepted
Expectation of the absolute value of the product of correlated jointly gaussians?
Probably not the best solution, but a direction that may be useful.
Suppose
\begin{align}
\begin{bmatrix}
X \\ Y
\end{bmatrix} \sim N_2\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix},
\begin{...
- 21.2k
3
votes
Accepted
Convergence in probability to a constant and absolute value (?)
$|E[X_n] - a| = o(1)$ tells you very little about the behaviour of the distribution of $X_n$ apart from its mean.
For example
if independent $Y_j \sim \mathcal N\left(\frac{1}{j^2}, j^2\right)$ and $...
- 42.1k
3
votes
Distribution of errors
This sounds like a nonlinear regression problem. Your notation for the problem essentially reverses many of the standard statistical conventions (e.g., you are using a hat for the true value rather ...
- 133k
3
votes
Accepted
Compounding a Gaussian distribution with variance distributed according to the absolute value of another Gaussian distribution
You can treat $y$ as $x z$, where $x$ and $z$ are standard Gaussian variables. Thus $y$ follows the Product Normal distribution.
- 563
3
votes
Accepted
Need help understanding how only variable A can be correlated to the absolute value of A-B
There are many natural ways this can occur. One is that even a single influential outlier can control the correlations. This situation will be obvious and scarcely needs explaining.
To find other ...
- 334k
3
votes
Why square the difference instead of taking the absolute value in standard deviation?
A different and perhaps more intuitive approach is when you think about linear regression vs. median regression.
Suppose our model is that $\mathbb{E}(y|x) = x\beta$. Then we find b by minimisize the ...
- 822
2
votes
Why square the difference instead of taking the absolute value in standard deviation?
This is an old thread, but most answers focus on analytical simplicity, which IMO is a weak argument in times of computers (although numerical stability might be an issue when using absolute values in ...
- 5,720
2
votes
Accepted
Expansion of $\mathbb{E}(|X|)$ and $\mathbb{E}(|X|+|Y|)$
In both cases, it is the first of the pair. You are considering random variables $|X|$ and $|Y|$, so you take them (not $X$ and $Y$) as basic units of analysis.
- 69.5k
1
vote
Validity of work-around to fit power curve for mixed-sign data
I do expect the same equation for positive or negative inputs, the only thing that should change is the sign of the output, which is easy enough to account for. This has been corroborated with other ...
- 11
1
vote
Why sum of squares to calculate dispersion
There are two broad reasons why sum-of-squares is generally used for measuring dispersion instead of sum-of-absolutes:
The sum-of-squares quantities (e.g., sample variance, etc.) turn out to have &...
- 133k
1
vote
Distribution of a spread of observations in triplicate sample taken from Gaussian distribution
If the question is about finding the distribution of the range of sample of 3 independent observations from a common normal distribution with known mean and variance, then numerical integration will ...
- 4,505
1
vote
Solve an inequality finding the upper bound
That integral function $T$ is an expression for the mean (or expectation value) of $Y$ when it follows the distribution $F$. See the bullet point 'Formulas in terms of CDF' on Wikipedia's article ...
- 86.5k
1
vote
MGF of the absolute Value of a Skellam RV
Let $Z$ follow a Skellam distribution with rates $\lambda_1$ and $\lambda_2$, and $K = |Z|$. The pmf for $K$ is:
$$p(k; \lambda_1, \lambda_2) = \begin{cases} e^{-\lambda_1 - \lambda_2} \left( \left(\...
- 530
1
vote
Need help understanding how only variable A can be correlated to the absolute value of A-B
I would say it is largely a mild curiosity, though there is a real effect
Note that the two pairs $(9,3)$ and $(1,1)$ give both the extreme values of Sample1 and the extremes of the absolute ...
- 42.1k
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