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7

Well, in Bayesian statistics, you don't just "make up" your priors. You should be building a prior that best captures your knowledge before seeing the data. Otherwise, why anyone should care about the output of your Bayesian analysis is very hard to justify. So while it's true that the practitioner has some sense of freedom in creating a prior, it should ...

6

Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. This condition is much stronger than what you actually need to get the result in the conjecture, so I will give you a ...

6

If you have a belief about the distribution of your data after seeing data, then why would you be estimating its parameters with data? You already have the parameters.

5

In case of many problems in statistics you have some data, let's denote it as $X$, and want to learn about some "parameter" $\theta$ of the distribution of the data, i.e. calculate the $\theta|X$ kind of things (conditional distribution, conditional expectation etc.). There are several ways how can this be achieved, including maximum likelihood, and without ...

3

The underlying statistical model is $X_1,\ldots,X_n\sim Bernoulli(p)$ iid ($X_i=1$ iff friend $i$ gets a ticket; $n=7$), i.e. $Y_n=\sum X_i\sim Binomial(n,p)$. You wish to estimate $p$, having observed $Y_n=0$ Here are two approaches. Rule of three The maximum likelihood estimate is indeed 0. You could construct a confidence interval. A rule of thumb in ...

2

I'm not sure how intuitive this is, but the main technical result underlying your statement of the Halmos-Savage Theorem is the following: Lemma. Let $\mu$ be a $\sigma$-finite measure on $(S, \mathcal{A})$. Suppose that $\aleph$ is a collection of measures on $(S, \mathcal{A})$ such that for every $\nu \in \aleph$, $\nu \ll \mu$. Then there exists ...

2

Under the assumption that $X$ and $Y$ are independent and continuous, \begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text{d}F_Y(y) \\&= \frac{1}{2} \int_{\Bbb R} \, \text{d}F_Y^2(y)\\&=\frac{1}{2}F_Y^2(\infty)-\frac{1}{2}F_Y^2(-\infty)\\&=1/2\...

2

It helps to go back to basic definitions. A probability space is the triple $(\Omega, \mathcal{F}, P)$ where $\Omega$ is the sample space, $\mathcal{F}$ is the set of subsets of $\Omega$, called "events", on which $P$ is defined (normally the Borel $\sigma$-algebra), and $P$ is a probability measure. A random variable is a function $f: \Omega \mapsto E$, ...

1

If there is a central interval where the density is $0,$ then the median of even a large number of observations cannot be anything close to normal. In the simulation below, the population distribution is a 50:50 mixture of $\mathsf{Unif}(0,1)$ and $\mathsf{Unif}(2,3),$ so that the density is $0$ in $(1,2).$ The simulation shows a histogram of 100,000 ...

1

This doesn't really seem like it should pose a problem. If you run a regression with "price" as the dependent variable, then there will be variation in price, even given all the independent variables. But that's common; I'd say it's almost universal.

1

You are doing nothing wrong with the test per se. Chi squared tests for stochastic independence. I believe you may have misunderstood what the p value and significance actually tell you about your data. Consider what the example you gave shows: If you know that someone is female, you would judge the chance of her liking tea to be higher than if the person ...

1

Provided the expectations exist, Cauchy-Schwarz inequality states $$E\left[(g(X))^2\right]E \left[(h(X))^2\right]\ge \left(E \left[g(X)h(X)\right]\right)^2$$ Choose $g(X)=I_{X>0}$, the indicator of the event $\{X>0\}$, and $h(X)=X$. And keep in mind that $$E(X)=E(X I_{X>0})+E(X I_{X<0})=E(X I_{X>0})$$

1

Adding to Tim's answer, this is more of a comment than a full answer, but I disagree with the premise that it is very important to know the underlying data distribution to develop a good model You say you have noticed this ...in many textbooks Can you provide some references to textbooks that make this assertion ? Having said that, let's not ...

1

First of all, you never "know" the distribution. You make assumptions about distributions. If people say things like "height follows normal distribution" they do not mean that there's some force in the universe that makes the height be exactly consistent with some made up mathematical function. They mean that they've chosen normal distribution as an ...

1

Partial advice, in no particular order, to succeed in your DS interview: Study through a book like Hastie et al.'s Elements of Statistical Learning or Barbers's Bayesian Reasoning and Machine Learning; forget videos for a long-term plan unless you decide to follow a long series of lectures. Do not get sucked into Deep Learning to begin with; it is cool and ...

1

When $z < 0$, instead of writing $$F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr( X \leq \frac{1}{z}) = F_X(\frac{1}{z})$$ you must write $$F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr(1 \geq Xz) = Pr( X \geq \frac{1}{z}) = 1-F_X(\frac{1}{z})$$ because every time you multiply or divide by a negative number (or in general any time you apply a nonincreasing function to ...

1

I find it difficult to follow your question because of the notation (for instance, what are "examples"? Do you mean data points?). Anyway, maybe the following example will clarify your question. The k-means method is iterative; let's assume we have reached iteration i and the results at this point look like that (point for data and crosses for cluster ...

1

Although the mean absolute error seems like a natural error measure at first sight, it has the drawback that it is not related to confidence intervals in a natural way. That's why the mean quadratic error (aka the variance) is used instead. Another interesting property of the variance is that the arithmetic mean minimizes the variance. The mean absolute ...

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