New answers tagged

2

Let's first explore how much progress we can make without trying to solve for the x's in terms of the y's and by avoiding a direct calculation of the Jacobian (according to the Principle of Mathematical Laziness). From $$\mathrm{d}y_1 = -e^{-x_1}\mathrm{d}x_1$$ and $$\mathrm{d}y_2 = -e^{-x_1x_2}\left(x_2\mathrm{d}x_1 + x_1\mathrm{d}x_2\right),$$ both ...


2

In your solution, you calculate the probability of getting tails twice as the square of the probability of getting tails on the first flip. This assumes that consecutive flips are independent. In fact, they are not, as the same coin is used throughout and so getting tails on the first throw means it is more likely to get tails on consecutive throws. This is ...


3

It looks like you're on the right track, but you're working way too hard on this one. Sooner or later algebraic and numerical mistakes will creep into even the best calculations. A good strategy is to minimize the amount calculation: the Principle of Mathematical Laziness. A key element of this principle is just-in-time computation: don't do any work ...


4

It's sometimes useful to recast a problem in terms that yield better search engine results. Here is an alternative formulation of your problem: We throw balls at random into $n=6$ urns, with equal probability. How many balls do we expect to throw until one urn contains $k$ balls? And there is actually a closed form solution to this question at Balls are ...


3

This is not a full answer, but it may be helpful. We can model your problem as an absorbing Markov Chain. The possible states are $n$-tuples of numbers between $0$ and $k$, $$\mathcal{S} := \{0,\dots,k\}^n, $$ each state marking how often each number between $1$ and $n$ has already come up. (Of course, $n=6$.) The transient states are those where all entries ...


3

Something that might be helpful in unifying these two views is the Hewitt-Savage(-de Finetti) representation theorem. The theorem says that $X_1,\dots,X_n$ are exchangeable precisely when they are independent and identically distributed conditional on some additional information. This is important in Bayesian statistics, because it means that an ...


2

Instead of focusing only on the distribution function, let's focus on equality in distribution. A finite sequence of random variables $X_1, \ldots, X_n$ is exchangeable if for every permutation $\pi$ we have $$ X_1, \ldots, X_N =_d X_{\pi(1)}, \ldots, X_{\pi(n)} $$ where $=_d$ means equality in distribution. Equality in distribution is equivalent to ...


1

You say that $X$ has some parametric distribution, which is a location-scale family. Then aspect of the distribution which is not influenced by the location or scale is said to measure some aspect of the distributions shape. By transforming $$ X_i \mapsto \frac{X_i-a}{b} $$ you changes $X$ to have some distribution from the same family, only with location ...


0

Maybe I'm misunderstanding the question but I fail to see why that expectation doesn't just blow up to infinity. Intuitively it is essentially just counting the number of times you draw red, given you have an infinite number of draws. And the only way this kind of infinite sum converges is if it is a geometric sequence inside some radius of convergence. But ...


1

Expand the covariance by using its definition: $$ \begin{align} \mathrm{cov}(X, Y) &= \mathrm{E}\left[(X-E[X])(Y-E[Y])\right]\\ & = \mathrm{E}\left[\left\{(-Y)-E[-Y])\right\}(Y-E[Y])\right]\\ & = -1 \cdot \mathrm{E}[(Y-E[Y])(Y-E[Y])]\\ & = -1 \cdot \mathrm{E}[(Y-E[Y])^2]\\ & = -1 \cdot \mathrm{Var}[Y] \end{align} $$ Also, notice that $$ \...


1

Weierstrass Approximation Theorem: Suppose $f$ is a continuous real-valued function defined on the real interval $[a,b]$. For every $\varepsilon > 0$ there exists a polynomial $p$ such that for all $x \in [a, b]$ we have $|f(x)−p(x)| < \varepsilon$ (or equivalently, the supremum norm $||f−p|| < \varepsilon$). The notes ask you to prove the result ...


0

Just to record a variant on the above calculation with a couple of shortcuts: From $\frac{x-\theta}{1+(x-\theta)^2}=\frac{x+\theta}{1+(x+\theta)^2}$ we see that the function $g(y)=\frac{y}{1+y^2}$ takes the same value at $y_1=x-\theta$ and $y_2=x+\theta$, hence so does $\frac{1}{g(y)}=y+\frac{1}{y}$. This function is 2 to 1 (except at $y=\pm 1$) so either $...


0

This follows my comment (using the approach shown in the Youtube video). Once you have the joint you can calculate the conditional. Sorry for the typo. Of course, $g_2(X,Y)=X$, and not $Y$.


0

The notation $$\mathbb E\left \{ \mathbb E\left \{ \mathbb E\left \{ Z \mid X , Y \right \} \right \} \right \}$$ is incorrecte, it should be $$\mathbb E [ \overbrace{\mathbb E \{ \underbrace{\mathbb E( Z \mid X , Y )}_\text{function of $(X,Y)$} \mid X \}}^\text{function of $X$}]$$ or $$\mathbb E [ \overbrace{\mathbb E \{ \underbrace{\mathbb E( Z \mid X , Y )...


0

You need to apply a functional derivative in this case, use the Euler-Lagrange formula: Given $$\mathcal{F}[f] = \int_a^b G(f(x), f'(x)) dx$$ $$\frac{\delta \mathcal{F}}{\delta f} = \frac{\partial G}{\partial f(x)} - \frac{d}{dx}\frac{\partial G}{\partial f'(x)}$$ Since the functional in your case does not depend on the derivative of $f$, the expression ...


1

If you have data $X_1, X_2, \dots, X_n$ randomly sampled from $\mathsf{Norm}(\mu, \sigma),$ with $\mu$ known and $\sigma$ unknown, then $V = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2$ has $E(V) = \sigma^2,$ with $\frac{nV}{\sigma} \sim \mathsf{Chisq}(\nu = n).$ The proof follows directly from the definition of $\mathsf{Chisq}(\nu = n)$ as the distribution of the ...


1

The differences between the discrete and continuous cases have been well explained in the comments. For the conditional expectation in the exponential case with density $f_X(x)=e^{-x/p}/p$ for $x>0$, I suggest first working out the conditional density as $$f_{X|Z=1}(x) = f_{X|X<Y}(x) = \frac{2}{p}e^{-2x/p},\quad x>0.$$ This utilizes $P(X<Y)=1/2$,...


0

Assume $x\ge 0$ so that $f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and $L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$ This function is clearly decreasing in $\theta$ so the smallest possible value for $\theta$ maximizes the likelihood. The largest observation is not allowed to exceed $\theta$ because ...


-1

${S}_{1}^2=0.8\times 0.2={S}_{2}^2=0.16$ ${S^2}=0.5\times 0.5=0.25$ $W_{1}=W_{2}=0.5$


4

See also this question; the proof is sketched in the related comments by @cardinal. Without loss of generality we can assume that the interval is $[0, \,1]$. Consider the following Bernstein's polynomial $$ B_n(x) := \sum_{k= 0}^n f(k/n) { n \choose k} x^k (1 - x)^{n-k} $$ which will provide an approximation of $f(x)$: we can prove that $B_n(x)$ tends to $f(...


0

I think you might be missing the fact that the expected value applies only when you do the sampling over and over, not on any one sample. On any given sample, a bad estimator (not unbiased, not MLE, high variance) might have a really good estimate, but we have no way of knowing by how much any given estimator misses, so we prove properties and take our ...


2

The result is correct, but the reasoning is somewhat inaccurate. You need to keep track of the property that the density is zero outside $[0,\theta]$. This implies that the likelihood is zero to the left of the sample maximum, and jumps to $\theta^n$ in the maximum. It indeed decreases afterwards, so that the maximum is the MLE. This also entails that the ...


2

As @whuber mentioned, you have $U=Y'AY$ where $A=\Sigma^{-1}-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\end{pmatrix}$. Note that $A$ is symmetric, so you can use the result in the question here to get the variance: $$\operatorname{Var}(U)=2\operatorname{tr}((A\Sigma)^2)$$ Also observe that $$A\Sigma=I_2-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\...


2

The joint distribution can be defined using the measure-theoretic definition: you have to know $\mathbb{P}((X,Y)\in A) $ for all $A\in \mathbb{R}\times \mathbb{N}$ for instance, and this works for any two random variables, you can extend this to any space of definition for $X$ and $Y$. On the other hand, to define a pdf you would need a reference measure. ...


0

@thomas-lumley I think that the Generalized Slutsky Theorem (Eugene Demidenko. Mixed models: theory and applications with R. John Wiley & Sons, 2013 page 662-663 - Theorem 48) gives general conditions which are also suitable for a wide range of stochastic functionals $q_{n}(\hat{\theta}_{n})$ where $\left\lbrace q_{n}\right\rbrace$ is a sequence of ...


0

x(4)2→4761. Nomalized feature →x−us where u is average of X and s=max−min=8836−4761=4075. Finally, 4761−6675.54075 = −0.47 So Answer is : −0.47


2

You haven't said how $h$ varies with $n$ or depends on the data. If $h$ is constant, then $\hat f_{n,h}$ will converge to something, but not to the true density of $X$. You also need to consider in what metric $\hat f_{n,h}$ converges in probability. Suppose $h$ is chosen so that the MISE of $\hat f_{n,h}$ converges to zero in probability, so $\hat f_{n,h}$ ...


1

You are on the right track. So starting from $$f_Y(y)=\int_0^{\infty} f(x,y)dx = \int_0^{\infty}\frac{\beta^\alpha}{\Gamma(\alpha)\Gamma(c)}y^{c-1} x^{\alpha+c - 1}e^{-(\beta+y) x} dx$$ $$= \frac{\beta^\alpha}{\Gamma(\alpha)\Gamma(c)}y^{c-1} \int_0^{\infty}x^{\alpha+c - 1}e^{-(\beta+y) x} dx$$ Now, this seems a known pdf, hence $$= \frac{\beta^\alpha}{\Gamma(...


1

In part a, it says uniformly distributed over the set $\{(u,v):0<u<v<1\}$, which means the region between lines $u=0, v=u,v=1$ ($v$ is in y-axis and $u$ is in x-axis). So, he joint PDF is 1/Area of this region. In part b, you can apply Jacobian technique. Another method is to calculate $F_{XY}(x,y)$ and differentiate wrt $u$ and $v$. It's a good ...


1

This webpage answers your question, http://nicolas-hug.com/blog/pdps. A minor issue is that I don't think the fast way always have the same result as the slow method (the definition). The fast way is more like averaging on the conditional distribution of the other covariates (not the strictly defined conditional distribution, but in the tree structure).


4

Hints: In effect, you will have $\sum\limits_{i=1}^n X_i$ ones out of $n$ with the rest zeros, and you want to know how many of these ones are in the first $m$ The conditional probability will not depend on $p$ and will end up being a hypergeometric distribution If you do not take that shortcut, you could investigate $\dfrac{P\left(\sum\limits_{i=1}^m X_i ...


0

You should only use differencing for removing stochastic trends (unit roots), not deterministic trends (linear, polynomial, seasonal or otherwise); beware of overdifferencing (see e.g. Bob Nau "Identifying the order of differencing in an ARIMA model". Other than that, ARIMA will fit your time series equally well (or equally poorly), just the $d$ ...


2

I assume the ANOVA model is $$Y_{ij}=\theta_i+\varepsilon_{ij}\quad,\small\,i=1,2,\ldots,k\,;\,j=1,2,\ldots,n_i$$ where $\varepsilon_{ij}$'s are i.i.d $N(0,\sigma^2)$ for all $i,j$. In other words, $Y_{ij}\sim N(\theta_i,\sigma^2)$ independently $\forall\, i,j$. Mean of the $i$th group is $$\overline {Y_{i\cdot}}=\frac1{n_i}\sum\limits_{j=1}^{n_i}Y_{ij}\quad,...


2

No, you can't find the value of a three-term expression (probability expression that includes three events) with only two-term expressions. You'd need additional conditional independence assumptions to reduce the three-term expressions into two-term ones.


2

I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$. Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $$f(x)=\frac{e^{-x}x^{p-1}}{\Gamma(p)}\mathbf1_{x>0}$$ with $p>0$ as in @Xi'an's answer. We have \begin{align} E\left[e^{t\ln Y}\right]&=E\left[Y^t\...


3

A picture of an event usually helps in finding its probability. Each panel graphs the relation $x/y \le z$ for typical values of the number $z \gt 0.$ (When $z\lt 0,$ no points in the unit square satisfy this inequality, so the relation is empty.) I have chosen units of measurement in which the constant $a$ is one unit. This does not affect the ratio $X/Y,...


2

The variance and mean of a Poisson distribution are equal, so $E[(x-\theta)^2]=\theta$ and $$E\left[\frac{(x-\theta)^2}{\theta^2}\right]=\theta/\theta^2=1/\theta$$


1

If some variables are truly binary, then if we adhere to the intuition that "nonparametric" is equivalent (in some sense) to 'distributional-assumption free,' then a linear regression is indeed nonparametric for binary random variables, because all binary rvs have distribution $\text{Bernoulli}(p)$ for some $p$, and so for a binary rv $X$, the ...


0

Consider the process X = P + Q. This is also a Poisson process with mean 8 (=3+5). Probability of a claim in this process belonging to P is 3/8 and to Q is 5/8. If we think of a claim belonging to P as a success, then the random variable Y denoting the number of successes (claims from P) out of the first 5 claims is a Binomial random variable with success ...


0

Welcome to Cross Validated. Since you marked your question as self-study, I won't spoon-feed you the full answer, but try to point you in the right direction. Look at the numbers you're entering for your null hypothesis probabilities. Do they look like probabilities to you? Look at where you got them from. What exactly does the table say those numbers ...


3

Cumulative distribution function of random variable is defined to be a function $F_X(x)=Pr(X \le x)$. It is a convention to use $\le$, though it is not a universal one. We use a subscript to denote which random variable are we referring to. The $x$ on the left, and the $x$ on the right are the same $x$. For example, if $X$ is the distribution of the height ...


1

A simple answer here. We construct random variable X representing correctness of model 1. X=1 if model 1 is correct, 0 otherwise. Similarly we define Y for model 2. Consider Pr(model 1 is correct and model 2 is correct|both are correct or both are wrong), since we know they give same prediction, thus we have this condition of both correct or both wrong. It's ...


1

A simple approach to this problem is considering it from a geometrical view of point. Firstly, we immediately know answer is within range [0.1, 1], then let's check if it's tight. Note regression is projection of $y$ on to column space of $x$. If vector $x_1$ and $x_2$ are almost perfect linear dependent, it's easy to see the projection of $y$ is almost the ...


0

I'm not sure a transformation approach using a Jacobian is the best way to do this. By independence, for the minimum we have $$P(W > t) = P(X > t, Y > t) = P(X > t)P(Y>t) = 1/t^2,$$ for $t > 1.$ So $F_W(t) = P(W \le t) = 1 - 1/t^2.$ Also, $$F_Z(t) =P(Z \le t) = P(X \le t, Y \le t) = P(X\le t)P(Y\le t) = (1-1/t)^2,$$ for $ t > 1.$ I will ...


0

When in doubt, deal with indicator functions (and other awkward creatures like absolute values) by breaking the problem into cases. Here we have $$F(x_1,x_2) = \begin{cases} (kx_1^2)(kx_2^2) & \text{if } x_1 \in (0,1) \text{ and } x_2 \in (0,1)\\ kx_1^2 & \text{if } x_1 \in (0,1) \text{ and } x_2 \gt 1 \\ kx_2^2 & \text{if } x_1 \gt 1 \text{ ...


0

Alternative way to describe the CDF without indicator functions It is common to split up the CDF into cases like for the uniform distribution between $a$ and $b$ you have: $$F_X(x) = \begin{cases} 0 &:& x < a \\ \frac{x-a}{b-a} &:& a\leq x\leq b\\ 1 &:& x>b \end{cases}$$ and the density is the derivative $f_X(x) = {F_X}^\prime(x)...


0

Well, I'll give it a try. There could be mistakes though I'm my logic. So total rides for a month are 10,000. On average there's a fault after 100 rides. So there'll be 10000/100 = 100 faults. Each fault costs 10€ to repair, so (10€)×(100 faults) = 1000€ Assuming all 300 bikes are in working condition, so 300 - 100 = 200 working bikes for the month. This is ...


3

On p. 2 they write that to use this model for regression they use squared loss, so no activation is needed, while for classification it is passed through softmax function and use cross-entropy loss.


3

If the target is not constrained, e.g. such as probability values, the final activation is typically dropped. As per with note (2) in the paper, they assume centred target values: Note that we omit the outer-most bias term as this is equivalent to centring the output


2

Some observations: The radius is $w$, but you're taking it as $w^2$ Don't forget the $|J|$ term (i.e. $rdrd\theta$) in the integral (or you can simply use area of the circle as well) The density will take two different functional forms for $[0,1]$ and $[1,\sqrt{2}]$. Visualise a growing circle inside the square $[-1,1]\times[-1,1]$.


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