New answers tagged mathematical-statistics
3
votes
Accepted
Difference between anova () and summary ()?
There's a lot going on here. First, you don't seem to be conducting a repeated-measures anova but a multilevel regression model, which you then test using anova functions.
Second, anova() and Anova(), ...
- 2,548
4
votes
What does it mean to say that a statistical model is a location family?
In order to have a formal treatment of location families of distributions, one must reckon with how a group of transformations acts on a random variable with certain distribution to generate a family ...
- 9,567
0
votes
What is the intuition behind the factorization theorem? (Sufficient statistics)
The factorization theorem states that $S(X)$ is sufficient if and only if the likelihood function of $\theta$ for data $X$ can be factored into the product of a function of $X$ (constant in $\theta$) ...
- 66
0
votes
Persistence of the sum of two AR(1) Processes
After reading Granger and Morris (1976) (thanks, @mlofton), it appears that $Y$ does not follow an AR(1) process, but instead follows an ARMA(2,1) process. Further, it takes some algebra, but I can ...
- 508
1
vote
Can we conclude joint convergence in distribution from marginal convergence in distribution for two (not independent) sequences of random variables?
A counterexample:
$X_n\sim N(0,1)$, $Y_n=(-1)^n(X_n)$. Then for even $n$, $(X_n,Y_n)$ converges in distribution to $(X,Y)$ with $Y=X\sim N(0,1)$ and for odd $n$ $(X_n,Y_n)$ converges in distribution ...
- 43.3k
1
vote
How can I prove that two algorithms for weighted sampling without replacement are equivalent?
Yes, the two schemes are exactly equivalent (ignoring computational costs and numerical stability concerns). We can prove this by showing that both schemes perform $m$ draws from the weighted ...
- 11
0
votes
How does the reparameterization trick for VAEs work and why is it important?
Lots of great answers already; I'd like to offer a high-level answer: reparameterization is useful when you want to express the gradient of an expectation as an expectation of a gradient. It is useful ...
- 2,490
4
votes
Accepted
Does $|X_n|+|Y_n|=o_p\left(1\right)$ imply $|X_n|=o_p\left(1\right)$?
It is correct.
$|X_n|+|Y_n|\geq |X_n|$, so for any $M$ and $\epsilon$,
$|X_n|+|Y_n|<M$ implies $|X_n|<M$ and so
$$P(|X_n|+|Y_n|<M)>1-\epsilon$$ implies
$$P(|X_n|<M)>1-\epsilon$$
...
- 43.3k
4
votes
Does the MLE converge in mean-square?
An example of an MLE that converges in probability but not in mean square is the ratio of two binomials. Let $X_n\sim Bin(n,p)$ and $Y_n\sim Bin(n,q)$, then the MLE of $(p,q)$ is $(\bar X_n, \bar Y_n)...
- 43.3k
0
votes
Any relation between two KL divergences?
This might better be done with a different method to gauge differences between distributions. As Kullback-Liebler (KL) is a divergence, the triangle inequality need not hold. Thus comparisons like you ...
- 97.5k
2
votes
Accepted
Finding the expectation of a categorical variable times a random amount
If you are just interested in setting up the integrals for the expectation, then maybe the following does that:
$$\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\frac{\sum _{i=1}^J e^{u_i} u_i}{\...
- 4,195
6
votes
Mathematical Theory of Monotone Transforms
It is often pointless to estimate such a transformation: extremely few statistical procedures require the underlying distribution to be Normal and any estimate will be imprecise anyway. But, as a ...
- 329k
1
vote
How is summation by parts technique used in this derivation?
Ok since whuber commented, I decided to look back into it and figured it out.
I think the following expression from wikipedia makes it easier to understand:
$$
\sum_{k=0}^n f_k g_k = f_0 \sum_{k=0}^n ...
- 33
Top 50 recent answers are included