Hot answers tagged

28

Certainly the mean plus one sd can exceed the largest observation. Consider the sample 1, 5, 5, 5 - it has mean 4 and standard deviation 2, so the mean + sd is 6, one more than the sample maximum. Here's the calculation in R: > x=c(1,5,5,5) > mean(x)+sd(x) [1] 6 It's a common occurrence. It tends to happen when there's a bunch of high values and ...


26

This is referred to as current status data. You get one cross sectional view of the data, and regarding the response, all you know is that at the observed age of each subject, the event (in your case: transitioning from A to B) has happened or not. This is a special case of interval censoring. To formally define it, let $T_i$ be the (unobserved) true event ...


10

Let $f_\sigma(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{x^2}{2\sigma^2}\right)$ be the Normal$(0,\sigma)$ PDF and $g(x) = \frac{1}{\pi}\left(1+x^2\right)^{-1}$ be the PDF of a Student t distribution with one d.f. Because the PDF of a Normal$(\mu,\sigma)$ variable $X$ is $f_\sigma(x-\mu) = f_\sigma(\mu-x)$ (by symmetry), the expectation equals $$\...


10

This is Samuelson's inequality and it needs the $\leq$ sign. If you take the Wikipedia version and rework it for the $n-1$ definition of $S,$ you will find that it becomes $${{ \left| X_i-\bar X \right| } \over S} \leq {{n-1} \over \sqrt{n}}$$


9

Here's an example that we can use to illustrate ui and ci, with some extraneous output removed for brevity. It's maximizing the log likelihood of a normal distribution. In the first part, we use the optim function with box constraints, and in the second part, we use the constrOptim function with its version of the same box constraints. # function to be ...


9

Although I'm not entirely certain of what your problem with linear regression is I'm right now finishing an article about how to analyze bounded outcomes. Since I'm not familiar with Beta regression perhaps someone else will answer that option. By your question I understand that you get predictions outside the boundaries. In this case I would go for ...


9

For any $n$ random variables $X_i$ , the best general bound is $\newcommand{\Var}{\mathrm{Var}}\Var(\max X_i) \le \sum_i \Var(X_i)$ as stated in the original question. Here is a proof sketch: If X,Y are IID then $E[(X-Y)^2] =2\Var(X)$. Given a vector of possibly dependent variables $(X_1,\ldots ,X_n)$, let $(Y_1,\ldots ,Y_n)$ be an independent vector with ...


9

Even without those simplifying assumptions, a bound can be obtained by combining a couple of usual tools: The variance of the difference of two correlated variables. It allows us to turn a two variables problem into an univariate problem. Chebyshev's inequality. It puts a bound on the probability of exceeding a given value. In some detail: $$\sigma^2_{X-Y}...


8

Literature: Most of the answer you need are certainly in the book by Lehman and Romano. The book by Ingster and Suslina treats more advanced topics and might give you additional answers. Answer: However, things are very simple: $L_1$ (or $TV$) is the "true" distance to be used. It is not convenient for formal computation (especially with product measures, ...


8

The hazard is indeed a rate. It is the expected number of events a person can expect per time unit conditional on being at risk, i.e. not having died before. Say we are studying the time until you get the flu [influenza] , and we measured time in months and we got a hazard rate of .10, that is, a person is expected to get the flu .10 times per month assuming ...


8

When a response or outcome $Y$ is bounded, various questions arise in fitting a model, including the following: Any model that could predict values for the response outside those bounds is in principle dubious. Hence a linear model might be problematic as there are no bounds on $\hat Y = Xb$ for predictors $X$ and coefficients $b$ whenever the $X$ are ...


7

I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent. The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion: \begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= E(X^2) \\ &= \text{Var}(X) + ...


6

I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is always less than $X$. The variance of $Z$ is clearly going to be larger in such a case than the variance of $X$. This is an example where $X$ and $Z$ are ...


6

Good bounds for the Poisson binomial distribution Consider the following lemma: Given $0 < p_1 \le \dots \le p_N < 1$ with $N \ge \frac{2}{p_1}$, define $\mu := \frac{1}{N}\sum_{k=1}^N p_k$. For $m \le N+1 -\frac{2}{p_1}$, the tail of the Poisson binomial distribution with parameters $N$, $\{p_k\}_{k=1}^N$ is bounded by the binomial distribution ...


6

I work in health services research. We collect patient-reported outcomes, e.g. physical function or depressive symptoms, and they are frequently scored in the format you mentioned: a 0 to N scale generated by summing up all the individual questions in the scale. The vast majority of the literature I've reviewed has just used a linear model (or a ...


5

Let $R_i$ be a dummy which equals one for respondents and zero for non-respondents, $Y_i$ the outcome and $D_i$ the treatment variable from your randomized experiment. You cannot observe the counterfactual quantities that you want to compare, i.e. $E[Y_{i1}|R_i = 0|D_i = 1]$ and $E[Y_{i0}|R_i = 0, D_i = 0]$, due to non-response but you know their probability ...


5

The general asymptotic result for the asymptotic distribution of the sample variance is (see this post) $$\sqrt n(\hat v - v) \xrightarrow{d} N\left(0,\mu_4 - v^2\right)$$ where here, I have used the notation $v\equiv \sigma^2$ to avoid later confusion with squares, and where $\mu_4 = \mathrm{E}\left((X_i -\mu)^4\right)$. Therefore by the continuous ...


5

This is an idea how to solve it which uses the identity $$\frac{1}{S}=\int_0^\infty \exp(-tS)dt$$ which was proposed by Did here. You could use \begin{align}E\left(\frac{1}{x^2+1}\right) &= \frac{1}{\sqrt{2\pi}}\int_0^\infty \int_{-\infty}^{\infty}\exp\left(-t(x^2+1)\right)\exp\left(-\frac{x^2}{2}\right)\mathrm dx \mathrm dt\\ &=\int_0^\infty \exp\...


5

There is no upper bound. Intuitively, if $X$ has substantial support along a sequence approaching $1$, then $1/(1-X)$ could have a divergent (arbitrarily large) expectation. To show there is no upper bound, all we have to do is find a combination of support and probabilities that achieves the desired expectation of $a$. The following explicitly constructs ...


5

After simplifying the problem by means of routine procedures, it can be solved by converting it into a dual minimization program which has a well-known answer with an elementary proof. Perhaps this dualization is the "subtle step" referred to in the question. The inequality can also be established in a purely mechanical manner by maximizing $|T_i|$ via ...


5

You need to use the formula $$ \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}} \Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y)) $$ where $\Var $ is the variance operator. Take it term for term, write $\mu=\E X, \sigma^2=\Var X$, $\E (XY | Y= y) = \E (yX) = y \E (X) =\mu y$ with variance (over $Y$) $\Var (\mu Y) $ which is ...


5

The condition on the norm $$||\Sigma(\theta) - I_{p}||_{1} = ||\Sigma(\theta) - I_{p}||_{\infty} \leq a$$ implies that $$|\sigma_{ii} - 1| + \sum_{j\neq i}|\sigma_{ij}| \leq a \quad \forall i.\qquad (1)$$ If $\sigma_{ii} \geq 1$, then (1) implies $\sum_{j\neq i}|\sigma_{ij}| \leq a$, and therefore: $$\sigma_{ii} - \sum_{j\neq i}|\sigma_{ij}| \geq 1 - a.\...


5

It is immediate (from the definitions) that the lines must pass through extremal points of the point set. Because at least one of them must contain at least two of the points (provided there is more than one point!), at least one of them will be determined by an edge of its convex hull. Thus, for $n$ points with $m$ extremals, there is an algorithm ...


4

Let's call our random variables $X_1$, ... $X_n$. I assume that you meant that they are integer valued, also look at the comments to your original question. Let's look at the case that there is a unique minimum which is $j$ and is assumed by $X_i$ (and only $X_i$). The probability for this case is $$p(X_i = j) \prod_{k \neq i} p(X_k > j) = (\frac{1}{x})(\...


4

A simple way: Not having a distribution for $X$ makes it a bit tricky, but at least in the case of normal errors, I think this can be done. I think there's no useful upper limit on $c$ provided by the data in the scenario you describe, since the unobserved values (the ones without error) might have any distribution - as such very large values of $c$ are ...


4

For the general case, the answer is no. For the specific cases, it is also no. A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ var (x)=a^{-1}$ . Take $ x $ and $ y $ as independent, and we have: $$ var (z) =E [var (z|y)]+var [E (z|y)]=E [y^2a^{-1}]+var [y]=var (y) + E (y^2)a^{-1}=\...


4

Clearly not. An easy counterexample (here done in R), that I think satisfies all your constraints: set.seed(239843) x=rnorm(100,100,1) y=rep(c(0.01,0.99),times=50) z=x*y var(x) [1] 0.8413043 var(y) [1] 0.2425253 var(z) [1] 2425.296 What's going on: x is a series with mean 100 and sd 1. y alternates between 0.01 and 0.99. z=xy therefore alternates ...


4

Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series. Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X_2, \ldots, X_N)$ is $$\mu_k(\mathrm{X}) = (X_1^k+X_2^k+\cdots+X_N^k)/N,$$ which is a random variable, and the variance is $$\text{var}(\mathrm{X}) = \mu_2(\...


4

Per Chebyshev's inequality, less than k -2 points can be more than k standard deviations away. So, for k=1 that means less than 100% of your samples can be more than one standard deviation away. It's more interesting to look at the low bound. Your professor should be more surprised there are points which are about 2.5 standard deviations below mean. But we ...


4

Use the following identity: $_1F_1(x, y, t) = e^{t} {}_1F_1(y - x, y, -t)$. Thus the inequality $$_1F_1(a, a + b, -t) \geq {}_1F_1(ca, ca + cb, -t),~c > 1$$ can be rephrased as $$\begin{aligned}&e^{-t} {}_1F_1(b, a+b, t) \geq e^{-t} {}_1F_1(cb, ca+cb, t) \implies \\ &{}_1F_1(b, a+b, t) - {}_1F_1(cb, ca+cb, t) \geq 0\end{aligned}$$ The original ...


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