#StackBounty: #probability #sampling #multinomial Distribution of number of objects in Simple Random Sampling with Replacement (SRSWR)

Bounty: 50

Suppose we have an urn containing $m in mathbb{N}$ objects and we sample with replacement $n in mathbb{N}$ times with equal chance of sampling any object in any draw. Let $1 leqslant K leqslant min(n,m)$ be the number of different objects we sample. We can represent this problem as follows:

$$mathbf{N} sim text{Multinomial}(n, (tfrac{1}{m}, cdots, tfrac{1}{m})) quad quad quad quad K equiv sum_{i=1}^m mathbb{I}(N_i >0).$$

Find the probability mass function for $K$.


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#StackBounty: #probability #intuition What are some convincing arguments against marginalization property with probabilistic opinion po…

Bounty: 100

Suppose that we have four possible events $e_0, e_1, e_2 $ and $e_3$. These event are exclusive $e_i cap e_j = emptyset$ and exhaustive $cup e_i = Omega$. Now we ask two experts on their opinion as to the probabilities of the individual $e_i$. We get the results in the following table:

enter image description here

There are numerous ways to combine their opinion to get a resulting probability distribution. There are two papers in particular that provide a very good overview of the entire topic:

Genest, C., & Zidek, J. V. (1986). Combining Probability
Distributions: A Critique and an Annotated Bibliography. Statistical
Science, 1(1), 114–135. https://doi.org/10.1214/ss/1177013825

Dietrich, F., & List, C. (2017). Probabilistic Opinion Pooling. (A.
Hájek & C. Hitchcock, Eds.), Social Choice and Welfare (Vol. 1).
Oxford University Press.
https://doi.org/10.1093/oxfordhb/9780199607617.013.37

Some of the more prominent methods are:

  1. linear: take a weighted arithmetic mean of probabilities for each event

  2. geometric: take a weighted geometric mean of probabilities for each event and then normalize

  3. multiplicative: take the product of probabilities for each event and then normalize

The results of these methods are also included in the table. Each method can be justified based on firm mathematical results.

Now here is the catch. Suppose that we are no longer interested in the individual probabilities of $e_2$ and $e_3$, but we would just like to know what is the probability of $e_2 cup e_3$. There are two ways in which we can proceed:

  1. We can sum the resulting combined probabilities for $e_1$ and $e_2$.
  2. We can sum the experts’ opinions for $e_1$ and $e_2$ and then combine the probabilities.

As you see in the table, these two approaches do not yield the same results.
For example, for the geometric combinator method 1 yields 0.71 and method 2 yields 0.712.

If the two methods would be guaranteed to produce the same result, we would call this the marginalization property.
Lindley in his paper ”Reconciliation of discrete probability distributions” (sorry, I cannot find the exact citation) tries to argue that the entire concept of marginalization property is flawed, since in the first case, we are given more information by the experts. He also gives a numerical explanation for his supra-Bayesian model, but I find it very hard to follow. Furthermore, I am not very convinced by his explanation of receiving “more information”. His numeric results hold, but I have the feeling that this is due to the fact that his model is much more complicated and must account for the individual correlations between his experts.

Are there any intuitive arguments that would reconcile the marginalization property with the simple geometric and multiplicative operators?


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#StackBounty: #probability #distributions #estimation The distribution of STD/MAD for a Student-t

Bounty: 50

Where $X sim a$ symmetric Student-t Distribution $t_alpha$, with power law tail $alpha>2$, looking for the distribution of

$$ frac{sqrt{ sum_{i=1}^n x_i^2 }}{sum_{i=1}^n |x_i|}, $$

in order to estimate the properties of the statistic STD/MAD for a sample size $n$.

The problem is that both measures, MAD and STD will be correlated for a given draw, so we cannot work with two different distributions.


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#StackBounty: #probability #estimation The distribution of STD/MAD for a Student

Bounty: 50

Where X ~ a symmetric Student T Distribution $t_alpha$, with power law tail $alpha>2$, looking for the distribution of $frac{(sum_{i=1}^n x_i ^2 )^{½}}{sum_{i=1}^n |x_i|}$, in order to estimate the properties of the statistic STD/MAD for a sample size $n$.

The problem is that both measures, MAD and STD will be correlated for a given draw, so we cannot work with two different distributions.


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#StackBounty: #probability #incidence-rate-ratio Calculate probability of disease appearance

Bounty: 50

I am a doctor so please be kind with me and my basic understanding of statistics.

I have a dataset consisting of patients and their visits and I have labelled the presence of a specific kind of mole in their left and/or right hand with {0,1} values (0 = not present and 1 = present). The dataset looks like this:

       RH   LH
A1-001  0   0
A1-001  0   1
A1-001  0   1
A1-001  0   1
A1-001  0   1
A1-001  0   1
A1-002  0   0
A1-002  0   0
A1-003  0   0
A1-003  0   0
A1-003  0   0
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and so on…

So, that means that patient A1-001 had 6 visits with no presence of mole in his right hand during all visits and present of mole in his left hand in all visits except the first one.

I am interested in finding the probability of a hand developing a mole among only the patients that developed a mole in one hand and finding the probability of developing a mole in the other hand (given that the patient had already a mole in the other hand).

Furthermore, I want to know what is the probability of developing a mole within visits among the patients that developed a mole at some point in both hands

Could you help me model these simple questions?


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#StackBounty: #probability #bayesian #kernel-smoothing #maximum-entropy #multiarmed-bandit Thompson sampling with adaptive kernel densi…

Bounty: 50

This is an extension to this question, which is about handling arbitrary (potentially unbounded) reward distributions for the multi-armed bandit problem. Given a sequence of observed rewards $r_t in mathbb{R}$ for arm $i$, one could try to approximate the true reward distribution $R_i$ using a Gaussian posterior. It also occurred to me that one could use adaptive kernel density estimation to approximate the true reward distribution. Has the application of adaptive KDE to Thompson sampling been studied? How does it compare empirically with the Gaussian approach, or other possible approaches?


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#StackBounty: #probability #classification #ensemble #feature-weighting How to describe most important features of ensemble model as li…

Bounty: 100

I have created 3 different models and output of them is a class probability in binary classification problem. Models are bit different, showing importance from different features. I have of course one data matrix as a source for this exercise where 70% of data is used as training sample.

How one can summarize importance of different feature values to the final class prob estimate if only data matrix and list of features used is know besides this class probability estimate?

Individual models can be of course explained by different methods, but how one can explain avg ensemble predictions?


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#StackBounty: #probability #statistical-significance #mathematical-statistics #experiment-design Significance level example from "…

Bounty: 50

The paper reads:

Fisher immediately realized that this argument fails because every
possible result with the 6 pairs has probability (1/2)^6 = 1/64, so
every result is significant at 5%. Fisher avoided this absurdity by
saying that any outcome with just I W and 5 R’s, no matter where that
W occurred, is equally suggestive of discriminatory powers and so
should be included. There are 6 such possibilities, including the
actual outcome, so the relevant probability for (a) above is 6(1/2)^6
= 6/64 = .094, so now the result is not significant at 5%.

I do not understand how 1/64 is significant at 5% but 6/64 is not. It makes more sense to me that bigger of two numbers would be deemed significant as it describes something that happens more often.

What is wrong with my reasoning?


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#StackBounty: #probability #distance-functions #metric #measure-theory When does the Wasserstein metric attain inequality WLOG?

Bounty: 50

I’m reading a classic paper [1] that describes a version of the Wasserstein metric (aka Mallows metric), defined as follows. Let $F$ and $G$ be probabilities in $mathbb{R} ^B$, and let $U sim F$ and $V sim G$ be $B$-valued RVs with marginal distributions $F$ and $G$ and an arbitrary joint distribution. Then:

$$d_p(F,G) := inflimits_{ substack{U sim F \ V sim G} } EBig[ || U-V||^p Big]^{1/p}$$

The paper says the infimum is always attained for some joint distribution of $(U,V)$ and then goes on to prove various results by considering only the case in which the infimum is attained, “without loss of generality”. For example (Lemma 8.6):

In view of Lemma 8.1 [stating that the infimum is attained and that $d_2(.,.)$ is a metric], assume without loss of generality that $(U, V)$ are independent and $$d_p(F, G) = EBig[ || U-V||^p Big]^{1/p}$$

First, I don’t understand why this holds WLOG. Second, I’m confused about why the infimum would be attained for independent $(U,V)$: intuitively, wouldn’t opposite be true? I would expect the “distance” to be smallest when $(U,V)$ are as correlated as their marginals would allow.

References

[1] Bickel, P. J., & Freedman, D. A. (1981). Some asymptotic theory for the bootstrap. The Annals of Statistics, 1196-1217.


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#StackBounty: #probability #distributions #pdf #quantiles #cdf Find the PDF from quantiles

Bounty: 50

I have been presented a problem of this kind:
suppose I know the values of k quantiles for a continuous random variable $X$

$$X_{1%} = x_1, X_{5%} = x_2, dots , X_{99%} = x_{k}$$

so that

$$ F_X(x_1)=1%, F_X(x_2)=5%, dots, F_X(x_k)=99% $$

From these informations I want to draw the chart of the PDF.

I thought that I could proceed this way:

  • interpolate $F_X(x)$ to get a smooth CDF (for instance spline interpolation)
  • find the derivative (numerical) of the smoothed CDF at some points to obtain the PDF.

Are there other more direct methods to address this problem? Do you think my solution is solid?

Thank you.


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