# Tag Info

81

To me "1 in 80 deaths..." is by far the clearer statement. The denominator in your "1 in 80" is the set of all death events and that statement makes it explicit. There's ambiguity in the "1 in 80 people..." formulation. You really mean "1 in 80 people who dies..." but the statement can just as easily be interpreted as "1 in 80 people now alive..." or ...

67

First of all, my first intuitive thought was: "S2 can only be the same as S1 if the traffic death rate stays constant, possibly over decades" - which certainly wouldn't have been a good assumption in the last so many decades. This already hints that one difficulty lies with implicit/unspoken temporal assumptions. I'd say your statements have the form 1 in ...

43

It depends on whether you are describing or predicting. "1 in 80 people will die in a car accident" is a prediction. Of all the people alive today, some time within their remaining lifetime, one in 80 will die that way. "1 in 80 deaths are caused by a car accident" is a description. Of all the people who died in a given period (e.g. the time span of a ...

21

The two statements are different because of sampling bias, because car accidents are more likely to occur when people are young. Let's make this more concrete by positing an unrealistic scenario. Consider the two statements: One half of all deaths are caused by a car accident. One half of all people alive today will die in a car accident. We will show ...

12

Is my default interpretation indeed equivalent to Statement One? No. Let's say we have 800 people. 400 died: 5 from a car crash, the other 395 forgot to breathe. S1 is now true: 5/400=1/80. S2 is false: 5/800!=1/80. The problem is that technically S2 is ambiguous because it doesn't specify how many deaths there were in total, while S1 does. Alternately, ...

11

Here is my basic(!) explanation. You are totally correct that the loss function can be used to calculate how good (or bad) your algorithm is. However, it does so by using the known data set, or the data used to train the algorithm. Risk is the expectation of the loss function. This expectation applies to all input data, not just the data that you used. ...

11

To quote from my book, The Bayesian Choice (2007, Section 2.3, pp. 63-64) The Bayesian approach to decision theory integrates on the parameter space $\Theta$, since $\theta$ is unknown, instead of integrating on the sampling space ${\cal X}$, as $x$ is observed. It relies on the posterior expected loss \begin{eqnarray*} \rho(\pi,d|x) & = & \mathbb{...

8

Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density. Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\theta(x))dx + \int q(x) \log(q(x)) dx$$ The first term is the Cross Entropy $H(q, f_\theta)$ and the second term is the (differential) entropy $H(q)$. Note that ...

7

I would agree that your interpretation of the second statement is consistent with the first statement. I would also agree that it's a perfectly reasonable interpretation of the second statement. That being said, the second statement is much more ambiguous. The second statement can also be interpreted as: Given a sample of individuals in a recent car ...

6

The Bayes risk is the frequentist risk averaged over the parameter space against to the prior distribution $\lambda$. The notion turns a function of $\theta$, $R(\theta,\delta)$, into a positive number, $R(\delta,\lambda)$, and hence allows for a total ordering of estimators $\delta$, hence for the definition of the Bayes estimator$$\delta^\lambda=\arg\min_\... 6 It's a pretty general question, I'll try to lay out the main ideas in a simple manner. There are a lot of good resources which you can use for further reading, one which I can recommend is Shai Shalev-Schwarz "Understanding Machine Learning" which focuses on the theoretical foundations for machine learning. Put very simply, the idea in machine learning is ... 6 The statement as reported is wrong: A most standard example is provided by the James-Stein estimator: given X\sim\mathcal{N}_p(\theta,I_p) (p>2), assuming \theta is estimated under the square error loss,$$L(\theta,\delta)=||\theta-\delta||^2$$the estimators$$\delta_0(x)=x\qquad\text{and}\qquad \delta_{2(p-2)}(x)=\left(1-\frac{2(p-2)}{||x||^2}\right) ...

5

Bayes risk is a framework for making decisions. You identify a statistical model for your data and incorporate prior knowledge to obtain a posterior distribution via Bayes rule. To make a decision, you decide on a loss function L associated with a set of possible decisions. Your optimal decision, is the one that minimizes your Bayes risk, i.e. minimizes the ...

5

Because the main problem concerns applying a fully general and abstract formula to a somewhat complicated model (regression), let's address it by examining a simple concrete case. Ordinary regression is a good choice because it is well known, well understood, and serves as the archetype of all more complex regression models. But even this comes in several "...

5

The basic difference is that the two statements refer to different populations of humans, and different time frames. "One in 80 deaths is caused by a car accident" presumably refers to the proportion of deaths in some fairly limited time period (say one year). Since the proportion of the total population using cars, and the safety record of the cars, have ...

4

Your remark that $$R(\theta,\delta)=\int x L(\theta,\delta(x))f(x|\delta(x))\text{d}x$$ is doubly incorrect. It should be$$R(\theta,\delta)=\int \overbrace{L(\theta,\delta(x))}^{\text{no }x}\underbrace{f(x|\theta)}_{\theta\text{ not }\delta(x)}\text{d}x$$ In this special case of yours $X$ has a finite support $\{\theta-1,\theta+1\}$, the integral is ...

4

(I am assuming the loss function here is the squared error loss). You have the formula for Bayes risk right, but you have the second moment of the prior distribution wrong, and then algebra at the end wrong. I am going to proceed with the solution by first finding $R(\theta, \delta)$, and then $r(\delta, \pi)$, using the following two equations. $$R(\theta,... 4 You are of course right and it is a common mistake to describe an odds ratio like a relative risk ratio. I would suggest that it would be helpful to propose a more appropriate phrasing to them such as "suggesting that the odds of [participants] [doing X in] [A condition] is 1.90 times higher than in [B condition]." Once the authors realise that is all they ... 4 Suppose the dataset is \mathcal{D} = \{X_1, \dots, X_n\} where each data point X_i is drawn i.i.d. from some distribution f_X. The true risk is:$$R(h) = E_{X \sim f_X}[\mathcal{L}(X, h(X))]$$Show that E_{\mathcal{D}_n}[R_e(h)] = R(h) Start with the LHS:$$E_{\mathcal{D}_n}[R_e(h)]$$Plug in the expression for the empirical risk R_e(h):$$= ...

4

Age $\times$ risk factor interaction are, as mentioned, important. Some risk factors may have a different form of association in later life. Low values of some risk factors can be associated with mortality due to them being possible indicators of poor health in later life, leading to a u-shaped relationship. There is an interesting article by Abdelhafiz et ...

4

The answer is no. I'll show a simple counterexample where $p(x \mid y=0)$ and $p(x \mid y=1)$ have the same mean, but it's possible to construct a linear classifier with misclassification rate better than chance. (Side note: chance-level performance is only 0.5 when the marginal probabilities of each class are equal; if one class were more probable than the ...

4

First, denote by $g(x) = \mathbb{1}_{\eta(x) > 1/2}$ the Bayes optimal classifier, hence $$P(g_n(X) \neq Y) - P(g(X) \neq Y) = L(g_n) - L^* = 2\mathbb{E}|\eta(X) - 1/2|\mathbb{1}_{g_n(X) \neq g(X)},$$ this identity hold for any classifier $g_n$, in particular for plug-in classifier. Now let's notice that when $g_n(x) \neq g(x)$ it holds that $|\eta(x) - ... 4 In my understanding, the expected value of a random variable is not necessarily a good description of it. This depends on what you mean by "description". The expectation has a number of interpretations, all of which might or might not be "good" for you. In frequentist terms, it is the long-run average of a data-generating process. If you draw from a random ... 4 The following analysis will reveal just how little "uncertainty," measured in terms of variance (or anything related to it), is connected to Shannon entropy: the volatility may converge toward certainty while the entropy may grow without limit. This happens even when there is a vanishingly small probability that$X_n$may differ from the constant$...

4

Since $\hat f$ and $f^*$ are defined as risk minimisers, points i. and ii. are intimately connected. I think that this amount to a problem of uniform convergence of the risks over the class $F$. Moreover, you should be careful when defining the convergence type "$\to$" and what you are exactly looking for. A reference you may want to look at is ...

4

For risk ratios, the averaging is done on the log scale. So, if you take the log of y, then compute the weighted average, and then back-transform via exponentiation, you get the expected result (0.24). For example, in R: w <- c(4.3, 3.6, 4.1, 4.3, 3.7) y <- log(c(0.11, 0.24, 0.14, 0.12, 2.92)) exp(sum(w*y)/sum(w)) yields 0.2485498 (which is 0.25 when ...

4

How would you define "lowest loss possible"? We are talking here about the possible actions $a$ given the current state of environment $\theta$, those actions may lead to some rewards $r$ where each reward is characterized with a utility $U(r)$. The rewards that are possible, can be considered as a random variable $Z$, which has a distribution that ...

3

This is best done through simulation. See my MATLAB code example and explanation below: %% Get S&P 500 price series d=fetch(yahoo,'^GSPC','Adj Close','1-jan-2014','30-dec-2014'); n = 1; % # of shares p = d(end:-1:1,2); % share price, the dates are backwards PV0 = n*p(end); % portfolio value today %% r=price2ret(p,[],'Continuous'); % get the continous ...

3

Let's take the median, which is the 0.5 quantile, and is a kind of "typical middle value". The median is the value $Q_{.5}(Y)$ such that an equal number of observation are less than and greater than the value (for an odd sample size), or the average of the two central values (for an even sample size). For example, the median of $\{99,100,101\}$ is $100$. For ...

3

The risk in machine learning context is the expectation of the loss function over the random variable. Let's say you have a random variable $x$. If you apply the loss function $L(.)$ to this random variable you get another random variable $y=L(x)$. The expectation is not a function, it's rather an operator $E[.]$, when you apply it on the random variable $y$...

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