New answers tagged normal-distribution
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How to calculate the probability of this combination of normal distributions?
This is not an answer, but how I would think about this:
Let $Q,R,S,T$ be standard (multi)normal variables of dimensions $1,n-1,1,m$.
Let $e=\sigma a_1$, $f=\sigma a_{2..n}$, $g=\sigma b_{2..n}$ and $...
- 3,498
2
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Entropy of $\ell^p$ norm of multivariate Gaussian
A general result
The following requires only the most elementary calculation (a little arithmetic) and keeping track of some standard mathematical definitions. It generalizes the concept of a "...
- 313k
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How can we maintain asymptotic normality with slight change?
The question is not stated clearly. Here is what I understand, but not everybody will necessarily understand the same.
A. For result "1)", set $\gamma_n \equiv \bar \mu_n - \mu_n \implies \...
- 56.5k
1
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Accepted
How to test for normality on paired samples between two treatments when the number of observations per treatment is unequal?
The number of pokes in 1 hour is count data, therefore I would analyze it with a GLM for a distribution of the Poisson family. You have 2 measurements on the same mouse, therefore mouse identity ...
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How to test for normality on paired samples between two treatments when the number of observations per treatment is unequal?
There is no reason to expect normality. Think about using the rank difference test or its regression model generalization described here.
You have to acccount for a poke being a response variable ...
- 84k
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How can we maintain asymptotic normality with slight change?
I won't solve this for you, but the way I'd go on about it is that (just to use an example) I'd assume $a_n\not\to 1$ (or any other of the required conditions not fulfilled). This means that there is ...
- 19.4k
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Is it reasonable to use the Kolmogorov-Smirnov test to assess the normality of a random variable?
No one here has mentioned the Liliefors test. The Liliefors test is like the Kolmogorov–Smirnov test, comparing the empirical c.d.f. with the c.d.f. of the normal distribution with the same mean and ...
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Which is largest, of a bunch of normally distributed random variables?
We have
$$\begin{align}
\mathbb{P}(\mathcal{E}) &= \mathbb{P}(\bigcap_{i=1}^n \{X_0 \ge X_i \}) = \mathbb{P}(\bigcap_{i=1}^n \{X_0 -X_i \ge 0 \})
\end{align}$$
Let us denote $Z_i = X_0 - X_i$ ...
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How to calculate $\int_{-\infty}^{\infty} \Phi(y+c)^{k-1}d(\Phi(y))$?
This integral is equal to $\mathbb{E}(\Phi^{k-1}(Y+c))$ for $Y$ follows the distribution $\mathcal{N}(0,1)$.
By applying this answer with $a = 1, n = k-1$, we can have a closed-form expression for ...
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Expected value of softmax transformation of Gaussian random vector
Late to the party, but there's another (perhaps less complicated) avenue to calculate the value of $s(x)$, and that's taking the average of $s(x)$ over a very large number of runs.
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Accepted
Calculating the mean and sd of a lognormal distribution from the log mean and log 5th percentile
Let's call the (as yet unknown) mean and standard deviation of the underlying normal $\mu$ and $\sigma$, so $X \sim N(\mu,\sigma^2)$,
and the (known) mean of the lognormal $m_Y$ with $5$th percentile ...
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Data transformation to normal distribution in analysis of variance (ANOVA)
There are various tests of normality. I'm not sure I'd call them "rigorous" and I'd advise against their use in this context. In any case, ANOVA does not require normal data, one of its ...
- 105k
4
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Accepted
Poisson Regression and Linear Regression give the same error
Your expectation is wrong. It's all about evaluation metrics and loss functions:
Poisson regression minimizes the average unit Poisson deviance.
lm() minimizes the ...
- 11.5k
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What is the probability of $P(X>Y>0)$ where $X$ and $Y$ standard normal distribution with correlation $\rho$?
Given that the joint density of $(X, Y)$ is $f(x, y) = \frac{1}{2\pi\sqrt{1 - \rho^2}}\exp\left(-\frac{x^2 + y^2 - 2\rho xy}{2(1 - \rho^2)}\right)$, the probability of interest is
\begin{align}
P(X &...
- 14.6k
2
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Accepted
All uncorrelated marginals are independent: Only for joint Gaussian?
The answer is yes.
This is a consequence of the Darmois-Skitovich theorem.
It states that if $X_1,\ldots,X_p$ are independent and $\alpha_i,\beta_i\neq 0$ for all $i\in\{1,\ldots,p\}$, then $\alpha^TX$...
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Accepted
Distribution of a sum of linear combinations of random variables, each drawn from a set of random variables
First, note that
$$
\begin{align}
Z&=\sum_{k=1}^m Y_k
\\&= \sum_{k=1}^m\sum_{i=1}^n w_{i,k}X_i
\\&= \sum_{i=1}^n(\sum_{k=1}^m w_{i,k})X_i
\\&= \sum_{i=1}^n w_i X_i
\\&=\mathbf{w}^...
- 10.4k
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Transforming skewed distribution of dependent variable in linear regression?
Adding 1 before taking the log is not great because of the reasons identified in the comments.
However, I'm guessing your DV is measured per county or something like that. It's unlikely you have ...
- 105k
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All uncorrelated marginals are independent: Only for joint Gaussian?
I can't quite proof it, but i think that yes, $X$ has to be Gaussian.
Also as a point of nomenclature: I'd call $\alpha^TX$ a linear combination.
Now let's say $E[X] = \vec 0$ and $Cov(X) = \Sigma$, ...
- 1,847
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How to derive the combined frequency distribution from two independent normally distributed population?
If the normal distributions are identical you could use Ordered Statistics theory(see casella and berger section 5.4). But I think a better(correct?) approach here is from Multiple Random Variables(...
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Transform normal distribution to skewed distribution without changing its support
Despite the fancy jargon, the above answer is incorrect over the support mentioned, as neither function is normalizable if the support is x>1, so the functions do not technically define a ...
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Probability distribution of measurements and Parameters of measurements
Regarding the first question and first image:
In this graph(above) if systematic error is zero then average value will be the true value! How's that possible?
You are right that this isn't normally ...
- 19.4k
1
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Accepted
Conditioning on two normal variables $E[x|y=y_0,x\leq k]$ and obtaining an analytical expression
Hint:
First, we prove the vector $(x,y)$ follows a bivariate normal distribution. It suffices to notice that
$$
\begin{pmatrix}
x \\
y \\
\end{pmatrix} =
\begin{pmatrix}
...
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