15

We still care. However, a large part of statistics is now based on a data-driven approach where these concepts may not be essential or there are many other important concepts. With computation power and lots of data, a large body of statistics is devoted to provide models that solve specific problems (such as forecasting or classification) that can be ...


12

We do care but usually either the issue is taken care of, or we're not making a specific distributional assumption with which we could apply those considerations. Many of the usual estimators for commonly used parametric models are either fully efficient under the usual distributional assumptions for that model or asymptotically efficient under those model ...


6

The answer is in the negative. For any real number $a$ define the function $$f_a(x) = f(x-a).$$ It is clear that when $f$ is a distribution function, so is $f_a;$ that when $f$ is supported on the real line, so is $f_a;$ and that both $f$ and $f_a$ have equal entropy. For $a\ne 0$ it is impossible that $f=f_a,$ though, for if so, $f$ would be periodic ...


3

The value you get from $t$ table is for the upper tail. For the lower tail, it is $-1.708$ and since your value is to the left of it, you'll reject the null hypothesis.


2

If $X$ is independent of $Y$ and $X \sim Y$ then $X$ and $Y$ are exchangeable. That means that for any measurable function $f$, one has $f(X,Y) \sim f(Y,X)$, and consequently $E\bigl[f(X,Y)\bigr] = E\bigl[f(Y,X)\bigr]$ when this expectation exists. Apply this result to $f(x,y) = \frac{x}{x+y}$.


2

since $X, Y$ are iid then for joint distribution we have: $$\forall_{(x,y) \in \Omega} f_{XY}(x, y) = f_{XY}(y,x)$$ then: $$E[\frac{Y}{X+Y}] = \iint_{\Omega} \frac{y}{x+y}f_{XY}(x,y)dxdy=\iint_{\Omega} \frac{y}{x+y}f_{XY}(\underline{y,x})dxdy=^{Fubini Theorem}\iint_{\Omega} \frac{y}{x+y}f_{XY}(y,x)\underline{dydx}=^{x<->y}\iint_{\Omega} \frac{x}{x+y}...


2

You are correct; here's a similar way to show it. The half-normal prior on the standard deviation $\sigma$ (with scale $a$) is: $$p_\sigma(\sigma) = \frac{\sqrt{2}}{a \sqrt{\pi}} \exp \left( -\frac{\sigma^2}{2 a^2} \right)$$ The precision $\tau$ is related to the standard deviation as: $$\tau = g(\sigma) = \frac{1}{\sigma^2}$$ Since this is an invertible ...


2

1) $Y=F(X)$ is a transformation applied on $X$, just like $Y=X^2$ or $Y=\sqrt{X}$. So, $F(X)$ is not the PDF of $Y$, it's a transformation applied on $X$. The author chooses this transformation function to be the CDF of $X$ (for some reason of course, but that's not the question here). So, that is not the PDF of $Y$ (if it is, where is $y$ in it?). We just ...


2

It seems to me that the link you are missing here is to the probabilistic / information theory interpretation of VAEs. When the capacity of your networks is large enough you will reach a point where the solution with a larger latent space does not keep more information than a smaller one. This is possible in VAEs because they produce a noisy representation ...


2

As mentioned in Variational Autoencoder − Dimension of the latent space, there is a heuristic upper-bound for the latent variable dimension: the size of the training data. If you encoder is sufficiently powered (we assume it is anyways for VAEs), if your latent variable is $N$ dimensional and you have $N$ training samples, then your encoder can simply ...


1

Well, you have to decide which model you want to assume behind your discrete data-points! If you simply draw linear lines between your points, then averaging the discrete data-points is almost exactly the same as calculating the area/width. (because the 2 outer most data-points would have half weight) So it's the method of fitting that makes the ...


1

I am not sure if it is what you are looking for, but there is a beautiful literature that studies relation between stochastic variables without the second moment: On a new measure of covariation among stable variables Correlation in $L^p$ spaces


1

Consider two i.i.d. standard Cauchy variates $X$ and $Y$. Obviously the Pearson correlation between $X$ and $Y$, although it can be calculated on the basis of a sample, is not meaningful in this case. The Spearman and Kendall correlations, however, don't make any assumptions about existence of moments, and, as such, are perfectly valid measures of ...


1

This is the analogue when you have to sum over a continuum. If you were to do the sum over all of the integers, it makes sense to write a sum over the integers. You can write the infinite sum as a limit of a sequence of partial sums. Fourier series do this. $$\sum_{\mathbb{Z}}$$ However, it doesn’t make sense to talk about a sequence of partial sums for $\...


1

The parameter $\sigma$ is generally taken to be finite, so the outcome $1/\sigma = 0$ is not admissible. (If you take the limit to an infinite variance, your distribution approaches a uniform distribution on the reals, which is not a proper distribution, so you would need to think about what distribution you actually have in that case.) From the score ...


1

In the article, this equation is reached by $\nabla_\sigma\log L(\mathcal{X})=0$, where $L(\mathcal{X})$ is the likelihood of the data. You multiply each side by $\sigma$ assuming it's a real number satisfying $\sigma>0$. Actually, while taking derivatives, and taking logarithms (e.g. $\log \sigma^2$), you've already made the same assumption.


1

The weighted average and weighted median would each be appropriate in some situations, but you'd need assumptions about the distributions to get any confidence interval. As an example, suppose you want the median survival time among 30 people, where you know: The group of twenty men has median survival of $10$ months The group of ten women has median ...


1

I would say you can use anova for analyzing noise level measurements, see this stored google search which links many papers using anova in the analysis of noise measurements. As you say, noise measurements in decibel cannot really be added, say, if the problem is finding the resultant noise level from two simultaneous independent sources, like a jet taking ...


1

MLE of $\theta$ is more precisely given by $$\hat\theta_{MLE}=\overline XI_{0\le \overline X\le \frac12}+\frac12I_{\overline X>\frac12}=\begin{cases}\overline X&,\text{ if }0\le\overline X\le \frac12 \\ \frac12&,\text{ if }\overline X>\frac12 \end{cases}$$ Method of moments estimator of $\theta$ is as you say $\hat\theta_{MOM}=\overline X$. ...


1

If your sample is large and your data is independent and identically distributed, you may assume that all the assymptotic properties of the maximum likelihood estimator are valid: (1) Consistency; (2) Functional invariance; (3) Efficiency; (4) Assymptotic mean square error. See Wikipedia for details or Introduction to the Mathematical andStatistical ...


1

If given $X$; $Y$ and $Z$ are conditionally independent, we'll have $$E[Y|X=x,Z=z]=E[Y|X=x]$$ which will make $\gamma=0$ always.


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