# Tag Info

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One option is to plot additional curves from the green data. Here you plot one solid green curve connecting the mean values for each time to first detection. Instead, you could compute the 25th percentile of all the values at a given time to first decision. Repeat that for each time to first detection, and you'll get a whole curve of 25th percentile values. ...

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I highly recommend having a look at Caltech ML course by Yaser Abu-Mostafa, Lecture 8 (Bias-Variance Tradeoff) . Here are the outlines: Say you are trying to learn the sine function: Our training set consists of only 2 data points. Let's try to do it with two models, $h_0(x)=b$ and $h_1(x)=ax+b$: For $h_0(x)=b$, when we try with many different training ...

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The correct answer is $\frac{n-1}{n^2}S^2$. The solution is #4 here

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If the variances of two random variables are equal, that means on average, the values it can take, are spread out equally from their respective means.

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This problem of detecting a difference in two Normal distributions based on independent random samples from each was solved asymptotically by Pearson and Neyman in 1930 using the likelihood ratio test. (Specifically, the alternative hypothesis is that $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2.$ Under the null hypothesis, the likelihood ratio statistic is ...

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Assuming $X \sim {\cal N}(\mu, \sigma^2)$ then $$\Pr(X \leq m) = \Phi\left(\frac{m-\mu}{\sigma}\right)$$ where $\Phi$ is the cumulative distribution function of the standard normal distribution ${\cal N}(0,1)$. Thus, knowing $p=\Pr(X \leq m)$, one has $$\frac{m-\mu}{\sigma}=\Phi^{-1}(p)$$ and finally $$\boxed{\mu=m-\sigma\Phi^{-1}(p)}.$$ And you get ...

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My idea is similar to Nick Cox's comment above, but uses optimize in R, so you do not 'need' arithmetics (which of course should be preferred as it is exact). true_mean=5 #The unknown true mean var=1 #Your known variance t=3 #Your known cut off score t P<-1-pnorm(3,true_mean,var) #your known p-value opt<-function(x){(1-pnorm(t,x,var)-P)^2} #A loss ...

3

I would have said this as comment, except I don't have enough rep. As Glen_b already pointed out, the claim contains a typo. It should really be \begin{align}\sigma^2_M-\sigma^2_H&={1\over n} E(f^2)-{\theta^2\over n}-{\theta\over n} +{\theta^2\over n}\\&={1\over n}(E(f^2)-\theta)\\&={1\over n}(E(f^2)-E(f))\\&={1\over n}\int^1_0 ... 3 First, they compare crude (M) and hit-or-miss (H) there; they don't mention 'improvements' until after that section where the comparison is done. The page apparently contains a typographical error. The crude estimate (M) has a smaller variance, \sigma^2_M than the hit-and-miss estimate's (H) variance, \sigma^2_H. That is, ... 2 An alternative to using a Half-Cauchy distribution with a well-defined variance is a Half-Student-t with \nu>2 degrees of freedom, e.g. \nu=3.\pi(\nu)= \frac{12 \sqrt{3}}{\pi \left(x^2+3\right)^2},\,\,\, \nu>0.  This prior has semi-heavy tails and it should produce fairly similar results as the Half-Cauchy prior. You can visualise it in R ...

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In English it's saying that the mean squared error of the estimator is the sum of its variance and its squared bias. As the sample size increases the squared bias decreases more quickly (typically) than the variance, so eventually the mean squared error is nearly all variance and next to no bias. Why do we care? - we know then that the estimator converges ...

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This answer supposes that $X^TY$ (where $X$ and $Y$ are $n\times 1$ vectors) is a $1\times 1$ vector or scalar $\sum_i X_iY_i$ and so we need to consider the variance of a single random variable that is this sum of products. Since $X$ and $Y$ are independent random vectors, we note that $X_1, Y_1$ are independent random variables as are $X_2, Y_2$. Also, ...

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First of all, I think that you should look at the seasonal distributions separately, since the bimodal distribution is likely to be the outcome of two fairly separate processes. The two distributions might be controlled by different mechanisms, so that e.g. winter distributions could be more sensitive to yearly climate. If you want to look at population ...

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Are these distributions of something over time? Counts, perhaps? (If so then you might need something quite different from the discussions here so far) What you describe doesn't sound like it would be very well picked up as a difference in variance of the distributions. It sounds like you're describing something vaguely like this (ignore the numbers on the ...

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You can do this with an ANOVA analysis: my.locs$cluster = factor(rep(c(1, 2), each=5)) anova(lm(attribute ~ cluster, my.locs)) # Analysis of Variance Table # # Response: attribute # Df Sum Sq Mean Sq F value Pr(>F) # cluster 1 62.5 62.50 0.1109 0.7477 # Residuals 8 4510.0 563.75 This finds the variance within and ... 0 In the past I had a similar problem with the analysis of clinical data using 2-way ANOVA. What I did was to use the Anova function of the R package car and set the test type to III (type="III"). If you had posted more information, maybe I could provide also a code example. The results I got seemed reasonable. 1 (1) I don't think you've given an accurate example of bootstrap validation a) it does draw random sample with replacement, but I'm not sure it "ignores" anything b) it doesn't increase stability/accuracy of the model, but rather of the test error (2) This might be a useful starting point: Steyerberg: Internal validation of predictive models: efficiency of ... 7 Quick answer The reason is because, assuming the data are i.i.d. and$X_i\sim N(\mu,\sigma^2)$, and defining \begin{eqnarray*} \bar{X}&=&\sum^N \frac{X_i}{N}\\ S^2 &=& \sum^{N} \frac{(\bar{X}-X_i)^2}{N-1} \end{eqnarray*} when forming confidence intervals, the sampling distribution associated with the sample variance ($S^2$, remember, a ... 2 I didn't check that reference, but I guess they are assuming that$Y_i$'s are independent with$E(Y_i)=\mu$and$Var(Y_i)=\sigma^2$for$i=1,2,...,n$i.e. all the observation has the same (finite) mean$\mu$and (finite) variance$\sigma^2$. So first note that$E(Y_i^2)=Var(Y_i)+E^2(Y_i)=\sigma^2+\mu^2$. Also for$\bar{Y}=\dfrac{\sum_{i=1}^n Y_i}{n}\$ we ...

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Blockquote In general, I agree with the original hypotheses that higher-order terms are often associated with smaller variances. But, this also depends on the type of data. In plant breeding, a rule of thumb (Gauch, 1996, page 90) for multi-environment trials is that the variation in the data is: 70% location, 20% location-by-variety, 10% variety ...

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I have discovered that the regularity I described in my question has in fact been written about by several authors in the literature on Design of Experiments (DoE). It has been called the "hierarchical ordering principle" and also sometimes the "sparsity-of-effects principle." In the chapter on fractional factorial designs in Montgomery (2013, p. 290), he ...

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You are correct in assuming that you can't (shouldn't, really) analyse the data with the controls having zero variance. It sounds like you should consider using a two-way ANOVA on the raw data with the within day variance accounted for in the manner of a paired test. I wrote about the approach in this paper that is intended for pharmacologists with little ...

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