New answers tagged

2

Note, I am treating this as if it has been tagged with a self-study tag. Regarding (a), yes, the density must integrate to $1$, and it must be nonnegative everywhere. Integrate over that unit square and solve for $C$, or in other words, solve the following: $$ \int_0^1 \int_0^1 (2x^2y + Cy^5 )dx dy = 1. $$ Note that $\int_0^{\infty} \int_0^{\infty} f(x,y) ...


2

As you correctly realized no probability mass is placed outside $[0,1]^2$ by $f(x,y)$, so that restricting the integral bounds to $0$ and $1$ and setting the double integral equal to $1$ is indeed the correct approach. For (c) note that $\mathbb{P}(X\leq x, Y\leq y)=\displaystyle\int_0^x\int_0^y f(u,v)dudv$.


0

The standard error of the mean-percent difference (MPD) is quickly found by applictaion of the $\delta$-method: $f(x, y) = (y-x)/x$ so $\nabla f(x,y) = [y/x^2, 1/x]$. If the two samples are independent, $$\text{var}(MPD) = [K/A^2, 1/A]^T \left[ \begin{array}{cc} C & 0 \\ 0 & N \\ \end{array} \right] [K/A^2, 1/A] $$ Doing the algebra for SE gives ...


1

If $N$ is large enough, you can compute a confidence interval using the formula: $$ \frac{n_j}{N} \pm \frac{z}{N}\sqrt{\frac{(n_j)(N - n_j)}{N}} \\ $$ where $N$ is the total number of dice rolls, $n_j$ is the number of time that face $j$ appeared and $z$ is related with the confidence level. For a $95\%$ confidence interval, $z \approx 1.96$. Equivalently, ...


0

There is no standard way of choosing the better estimator in this case. The problem is know as the Bias-variance tradeoff. It depends on your problem and your preferences.


0

I would very seldom get away from the raw measurements. z-scores make the interpretation much trickier and have hidden assumptions. First, they assume that the standard deviation is the "gold standard" dispersion measure. This assumes among other things symmetry of the distribution. Second, some z-scores involve a standardization procedure where one ...


1

A good way to remedy your problem with the definition may be to think of the statement $\ P(X = a_j \ \text{for some} \ j ) = 1$ in "words" and apply this to your example. Suppose we define our list of outcomes (the $a_i$) by $a_1 = 0$, $a_2=1$, $a_3 = 2$ which is all possible numbers of heads. What the statement $\ P(X = a_j \ \text{for some} \ j ) = 1$ ...


1

You project on an infinite dimensional space spanned by $x_{t-1}, x_{t-2},\ldots$ The fact that a single $x_{t-k}$ for some $k$ is zero does not make such infinite dimensional space become the real line, it continues to be an infinite dimensional space.


2

In the case of paired observations you only fit one mean, in the case of independent samples you fit one mean for ech population.


1

We do make use of degrees of freedom while calculating test-statistics for independent samples t-test. We make use of the pooled std.dev in the test statistic - $S_p = \sqrt{\frac{(n_1-1)*(S_1)^2 + (n_2-1)*(S_2)^2}{n_1 + n_2 - 2}}$ where $S_1$ and $S_2$ are standard deviation of sample 1 and 2 respectively.


1

To be clear, you have 3 samples A,B,C of Chandler strawberries and you make 3 phenol determinations on each sample, altogether making a total of 9 phenol determinations on Chandler strawberries. (And similarly for Camarosa.) If that is true, then--generally inattentive, or not--your teacher is right that this is not a completely randomized design. The 3 ...


0

Regarding your last question, the expectation can be either w.r.t. $p(x,y)$ (the unconditional error) or w.r.t. $p(y\mid x)$ (the conditional error at each value $X = x$). Happily, minimizing the conditional error at each value $X = x$ also minimizes the unconditional error, so this is not a crucial distinction.


0

Reading the proof in more detail, it looks like the condition $\sigma_x^2 >0$ is required for the Wold decomposition theorem to apply. Above the theorem in the book this quantity is defined such that the situation you're describing falls under the case $\sigma_x^2 = 0.$


0

As @StephanKolassa says, this depends on your application.* One important thing to keep in mind: log transforms make a lot of sense when errors are proportional to the values, as this page discusses. If you are primarily interested in relative (or percentage) errors, staying in the log scale might be best. In your example, the difference between 10.1 and ...


2

Whether it makes more sense to evaluate your predictions on the original or on the log scale depends on your application, or on your loss function. Essentially: use a loss function that correctly measures how painful each loss is. (Which is two ways of saying the same thing.) We statisticians here can't tell you this. It depends on your domain. Note that ...


0

You can use a marcov chain. Here is a link. http://setosa.io/ev/markov-chains/


0

Yes, the calculation is correct. $$VIF = \frac{1}{1-R^2} \ .$$ If $VIF = 2$, then $1 - R^2 = 1/2$, so $R^2 = 1/2$. However, a pointer on the $VIF$: VIF of an independent feature $X_1 = \frac{1}{1-R^2}$, where $R^2$ is the coefficient of determination for the linear regression with $X_1$ as dependent and features $\{X_2, X_3, \ldots, X_n\}$ as ...


0

What you describe resembles the sum of absolute deviations (https://www.wikiwand.com/en/Average_absolute_deviation), except that you propose to add up absolute differences between pairs of points, while deviations are defined as differences from the mean. Both work, the main difference being perhaps that your proposition would produce more "dramatic" results,...


2

The definition you quote which is used with generalized linear models (glm) is not an exponential amily, it is an exponential dispersion family. For a fixed value of the dispersion parameter $\phi$ it is an exponential family (indexed by $\theta$), but when $\phi$ varies it is not. When used in glm's, the exponential dispersion family is used for ...


3

For the NFL theorem, you're right. I think it doesn't contradict and they mean something like this: if you design your ML algorithm for your specific task, it'll succeed on it, and possibly fail on many others; which is in parallel to what the original paper by Wolpert says according to wiki: "any two optimization algorithms are equivalent when their ...


2

Variance is the spread of the residuals: their tendency to depart from zero, as measured by their typical (root mean square) distance from zero. The left hand plot uses height to represent the residuals, but there is no discernible variation the amounts by which heights typically deviate from the zero value (near the middle) as one scans across the plot. ...


3

One value is collinear or constant. function_send is categorical with at least four levels AUDIT and FAS. Typically, model.matrix generates a covariate matrix for factors using dummy encoding, but first it uses droplevels to exclude redundant or unused factor levels, and only after complete case exclusion. That means that the level TAX is most likely ...


1

Although you might think of function_send as a single categorical covariate, it has three different levels that must be considered separately in Cox regression, or any regression for that matter. In this case the level TAX is taken as the reference level, and the values for the AUDIT and FAS levels represent the differences of each of those from that ...


0

The confidence intervals generated by OLS can be biased downwards in the heteroskedasticity case. So, we need to use robust standard errors if we are concerned about the t-stats of coefficient estimates. More information can be found here. The other problem is efficiency (ie. do we use the information embedded in our data well?). An observation with higher ...


10

Simple answer: if 2 variables are independent, then the population correlation is zero, whereas the sample correlation will typically be small, but non-zero. That is because the sample is not a perfect representation of the population. The larger the sample, the better it represents the population, so the smaller the correlation you'll have. For an ...


18

Comment on sample correlation. In comparing two small independent samples of the same size, the sample correlation is often noticeably different from $r = 0.$ [Nothing here contradicts @OmG's Answer (+1) on the population correlation $\rho.]$ Consider correlations between a million pairs of independent samples of size $n = 5$ from the exponential ...


2

With some simple algebra, you obtain the inverse relationship: $$\phi_k = -i \ln Z_k + i \ln A - 2 \pi B.$$ Hence, taking $A$ and $B$ to be constants (which should really be denoted as lower-case), you have: $$\begin{equation} \begin{aligned} \mathbb{E}(\phi_k) &= -i \mathbb{E}(\ln Z_k) + i \ln A - 2 \pi B, \\[10pt] \mathbb{V}(\phi_k) &= \mathbb{V}...


36

By the definition of the correlation coefficient, if two variables are independent their correlation is zero. So, it couldn't happen to have any correlation by accident! $$\rho_{X,Y}=\frac{\operatorname{E}[XY]-\operatorname{E}[X]\operatorname{E}[Y]}{\sqrt{\operatorname{E}[X^2]-[\operatorname{E}[X]]^2}~\sqrt{\operatorname{E}[Y^2]- [\operatorname{E}[Y]]^2}}$$ ...


1

Given that $\sigma_t^2$ is part of $I_{t-1}$ because of $$ \sigma_t^2 = \alpha_0 + \sum_{i=1}^p \alpha_i a_{t-i}^2 + \sum_{j=1}^q \beta_j \sigma^2_{t-j} $$ and given $$ \epsilon_t \stackrel{iid}{\sim} \text{WN}(0, 1), $$ couldn't you do $$ \begin{equation} \begin{aligned} E(a_t^2 | I_{t-1}) &= \text{var}(a_t | I_{t-1}) \\ &= \text{var}(\sigma_t \...


1

Suppose $A\in\mathbb R^{m\times n}$ and $B\in\mathbb R^{n\times m}.$ Then $\operatorname{tr}(AB) = \operatorname{tr}(BA).$ The proof of that is routine. So we have $Z\sim N_n(0, I_n)$ and $T\in\mathbb R^{n\times n}.$ Then \begin{align} & \operatorname E(Z'TZ) = \operatorname E(\operatorname{tr}(Z'TZ)) = \operatorname E(\operatorname{tr}(TZZ')) \\[8pt] = ...


6

A single random variable has a distribution; a sample mean from a random sample is a single random variable. Of course you can only observe its distribution by looking at multiple random samples (such as multiple sample means); then as the number of such samples increases the sample (empirical) cdf will approach the population distribution function. The ...


0

You can treat this as a two stage research project. First discover what transformations best fit the data, then use it. You will need to discover: Is it cyclic, with amplitude, frequency, and phase modulation, Multiple components? each with overtones? Scaling: linear, power,log, exponentiation? The following articles from the late 1960's and early 1970's ...


2

It is good that you have not succeeded, because the relation is not true: sometimes the variance of $\min(X,Y)$ exceeds the variance of $XY.$ Consider what happens to these two variances when any two random variables $X$ and $Y$ are simultaneously multiplied by a positive number $\lambda:$ because $XY$ is multiplied by $\lambda^2,$ its variance is ...


0

If we imagine that like WIFI access points (APs) transmitters and receivers can do same job and they are able to transmit and receive signal then, we can resemble your scenario to Localization Methods Based on Signal Lateration and Signal Strength Data Collection then those relevant papers which they used RSS-based localization techniques to evaluate their ...


3

It's hard to give a definitive answer, because "useful" and "useless" are not mathematical and in many situations subjective (in some others one could try to formalise usefulness, but such formalisations are then again open to discussion). Here are some thoughts. (a) Uniform convergence is clearly much stronger than pointwise convergence; with pointwise ...


0

The problem seems to be underspecified. As described, there might not even be chance involved and you do not mention any underlying distributions. Furthermore, it is not clear what kind of overall performance criteria are important. So here are some very generic/general ideas. You want to decouple the ranking from the scores, due to outliers. Furthermore ...


1

Standardization doesn't seem needed or especially useful here, as you still have to interpret the results using mean and SD (or whatever else is used to scale). Good old 16th century (well, 19th century graphically) logarithmic scale seems to work as well as anything: Graph drawn in Stata, so please don't ask me for R code.


0

First, to standardize your data you have to do it in two steps because you have two different populations. So first standardize your adult population and then the juvenile population. With this script, you can do it. It makes first an index of your data rows with the Adult population, and then take the rest. index <- runs$LifeStage == 'Adult' runs$...


0

The average of all results would be the "mean score" $\bar{F}(M)$ the problem you're describing with some data sets being outliers might be mitigated by using standard deviation (std) - $$\sigma(M)^2 = \sum(F(M,D)-\bar{F}(M))^2$$ Then you can use the mean divided by the standard deviation as a criterion for your methods, and rank them accordingly : $$C(M) = ...


0

There are several possible tests for this situation. A common choice is the two-sided two-sample Kolmogorov-Smirnov test. Note that this assumes that the distributions are continuous; if your variables are discrete this test tends to be quite conservative (lower than desired type I error, with corresponding reduction in power) As with the one-sample test,...


1

$f_X(x) = \tfrac{1}{B(2,2)}x(1-x)\mathbf{1}_{[0,1]}(x) = 6x(1-x)\mathbf{1}_{[0,1]}(x)$ et $\mathbb{P}[Y=y|X=x] = {15\choose y}x^y(1-x)^{15-y}$ Donc $$\mathbb{P}[Y=y] = \int \mathbb{P}[Y=y|X=x]f_X(x)dx = \int_0^1 {15\choose y}x^y(1-x)^{15-y}6x(1-x)dx$$ $$\mathbb{P}[Y=y] = 6{15\choose y}B(y+2,17-y)\int_0^1\tfrac{1}{B(y+2,17-y)}x^{y+1}(1-x)^{16-y}dx $$ $$\...


1

The property used here is that $\mathbb{Var}(\text{const}+X) = \mathbb{Var}(X)$. If you expand the terms out formally, from the second step to the third, you get: $$\begin{equation} \begin{aligned} \mathbb{Var} (\hat{\beta}_1) &= \mathbb{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1 x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right) \\[6pt] &...


1

If $X$ and $Y$ are independent random variables, then $$ \textrm{Var}(X+Y)=\textrm{Var} X+\textrm{Var} Y.\qquad (1) $$ This is because $$ \mathbb E\left[(X+Y)^2\right]=\mathbb E[X^2]+2\mathbb E[XY]+\mathbb E[Y^2]\overset{indep!}=\mathbb E[X^2]+2\mathbb EX\mathbb EY+\mathbb E[Y^2]=\left(\mathbb EX+\mathbb EY\right)^2. $$ Similarly if $a$ is deterministic and $...


1

What you wrote is not literally true, but it is correct in spirit. For given $n$ and $p$, your function $$ \binom{n}{k}p^k(1-p)^{n-k} $$ is the probability mass function of a random variable $S_n$ with mean $np$ and variance $np(1-p)$. In fact, this follows from the equality in distribution $$ S_n\overset{d}{=}X_1+\cdots+X_n, $$ (where $X_1,\ldots,X_n$ is a ...


0

@whuber's answer is right. I have thought a little bit more and have the following proof. As in the question, X,Y~iid and $-\infty<X,Y<+\infty$ and $U=\frac{X}{Y}$ , $V=Y$. So loosely speaking, $$ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)dydx = $$ $$ \lim_{\epsilon\to0}\bigg(\int_{-\infty}^{+\infty}\int_{-\infty}^{0-\epsilon}f(x,y)...


1

Define $$ v_t = r_{t}^2 - E_{t-1}[r_{t}^2] = r_{t}^2 - \sigma_{t}^2 $$ Plug this into the ARCH equation $$ r_t^2 - v_t = w + \sum_{i=1}^p \alpha_i r_{t-i}^2 $$ Rearranging yields the AR(p) model $$ r_t^2 = w + \sum_{i=1}^p \alpha_i r_{t-i}^2 + v_t $$ Yes, one should be able to estimate the model with least squares. However, there are some drawbacks ...


2

The inequality you have asserted is false: A simple counter-example is $X \sim \text{Bin}(2,\tfrac{1}{2})$ and $c=1$, which gives you the expectation: $$\mathbb{E}(\max(X,c)) = \frac{3}{4} \cdot 1 + \frac{1}{4} \cdot 2 = \frac{5}{4}.$$ For this counter-example we have: $$\frac{5}{4} = \mathbb{E}(\max(X,c)) > \max(\mathbb{E}(X),c) = 1.$$ There is a ...


4

Similar to winperikle's answer, just tightening the arguments a bit: $\max\{X, c\} \geq X$ and $\max\{X, c\} \geq c$. So, by taking expectation, $\text{E}\left(\max\{X, c\}\right) \geq \text{E} X$ and $\text{E}\left(\max\{X, c\}\right) \geq c$. Combining, we get $\text{E}\left(\max\{X, c\}\right) \geq \max \{\text{E} X, c\}$. These arguments can be ...


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