New answers tagged

0

You are right If n tends to large in binomial will tend to either normal distribution or Poisson. The Difference is in the Value of p . If p is close to 1/2 it will tend Normal and if p is very small and np < 5 or np <10 then it will tend to poison.


0

I think you have included an additional $y$ in the numerator and the index for $(1-y)$ seems to be not correct as well. $$f_Y(y|X=x) = \frac{y^x(1-y)^{1-x}}{\frac12}=2y^x(1-y)^{1-x}$$ Hence, $$f_Y(y|X=0) = 2(1-y)$$ which is nonnegative and integrates to $1$.


0

$X^3=[(X-\mu)+\mu]^3=(X-\mu)^3+3(X-\mu)^2.\mu+3(X-\mu).\mu^2+\mu^3$ Hence $E(X^3) = E(X-\mu)^3+3\mu E(X-\mu)^2+3\mu^2 E(X-\mu)+\mu^3=3\mu \sigma^2+\mu^3$. (The 1st term of the four is $0$ because of symmetry + finite support, the 3rd because $E(X)=\mu$) Now $\mu=7$ and $\sigma^2=\frac{35}{6}$, as you have already found, so you can get $E(X^3)$ directly ...


4

Make a table and apply the definitions. The expectation is found by multiplying each possible value by its chance and adding up those results. The residuals are the differences between each possible value and the expectation. The variance is the expectation of the squared residuals. The standard deviation is the square root of the variance. To illustrate,...


2

Your statement that "if there is only one ball left, then Pr(Xn = red) can either only be 1 or 0" isn't quite right. The value of Xn itself can only be 1 or 0, but the probability that Xn=1 is p, unless you have looked at all the other balls. When doing a sequential sampling without replacement, you do get more information about later draws, but that would ...


1

Note that $q(\cdot)$ is not indexed by $i$.


0

I think a more explicit answer can be given by looking at the value function definition. For a random walk (with no actions, a markov reward process), and no discounting, the value function for all states that do not transition into the success terminal state $s_{\text{uccess}}$ is given by $$V(s) = \underbrace{P(s_\text{failure})(0 + V(s_{\text{failure}}))}...


0

Here is a solution using the machinery of markov chains. The canonical form of a transition matrix with $r$ absorbing states and $t$ transient states is $$ \mathbb{P} = \pmatrix{\mathbf{Q} & \mathbf{R} \\ \mathbf{0} & \mathbf{I}} $$ where $\mathbf{I}$ is an $r\times r$ identity matrix, $\mathbf{0}$ is an $r\times t$ zero matrix, $\mathbf{R}$ is a ...


4

Letting $Y_i \equiv - 2 \ln R_i$ and noting that $R_i \sim \text{Rayleigh}(1)$ you have the CDF: $$\begin{equation} \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y_i \leqslant y) &= \mathbb{P}(-Y_i/2 \geqslant -y/2) \\[6pt] &= \mathbb{P}(e^{-Y_i/2} \geqslant e^{-y/2}) \\[6pt] &= \mathbb{P}(R_i \geqslant e^{- y/2}) \\[6pt] &= \exp \Big( -\frac{e^{...


7

Since $X$ is discrete, you can simplify a little: $$\lim_{n\to\infty}p(X_n=0) = \lim_{n\to\infty}\text{e}^{-{1 \over n}} = \text{e}^{\lim_{n\to\infty}{-{1\over n}}} = \text{e}^0=1$$ where we can go from the second to the third term by the continuity of the exponentiation function. The second statement follows from the first, as $n\cdot0 = 0$ and $n\cdot X ...


0

In Bayes classifier, you choose the class that gives maximum posterior probability/density. I'm not sure what you mean by Bayes classifier using ML. If you use ML, you won't take priors into account. For Bayes classifer, let $c$ be chosen class label for a given sample $x$: $$c=\operatorname{argmax}_c P(Y=c|X=x)=\operatorname{argmax}_c P(X=x|Y=c)P(Y=c)$$ If ...


2

Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question. If you are happy to drop either independence between $X$ and $Y$ or them having the same distribution, then there's hope. Same distribution. If you allow for dependent but identically distributed variables, ...


0

If you take the joint CDF over xy and derive it over just one of the variables - you're left with marginal PDF for that same variable. Let's prove using a simple joint distribution of two i.i.d. RVs X and Y ~Expo(1) On one hand, we can get to the marginal PDF through the joint PDF: $$ F_{XY}(x,y) = \iint_{XY} f_{XY}(x,y) dxdy = \int_{0}^{\infty }\int_{0}^...


1

The generic approach is to find the distribution of the rv $X-Y$, denoted $Z$, from the joint distribution of $(X,Y)$, which is a convolution exercise. Since the change of variables from $(X,Y)$ to $(Z=X-Y,Y)$ has Jacobian one $$\left|\dfrac{\text d(x,y)}{\text d(z,y)}\right|=\left|\dfrac{\text d(z+y,y)}{\text d(z,y)}\right|=\left|\begin{matrix}1 &1\\0 &...


6

Consider the following example: $$ X\sim\text{Unif}(0, 1) \\ Y = 1-X $$ $X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$. Note that this example relied on $X$ and $Y$ being dependent, identically distributed random variables. It is impossible for two independent, identically ...


1

You can simply find it via De-Morgan's Law. Let $A=T_1>t_1,B=T_2>t_2$ $$P(A\cap B)=1-P(A'\cup B')=1-(P(A')+P(B')-P(A'\cap B'))$$


2

You seem to substitute $P_1$ for $P(A|H_1)$, which is wrong. It's actually $P(A|H_1)\geq P_1$ because $A$ doesn't win only when the second toss is Heads, it can still win when the second toss is tails; an example case is tails, tails, heads, heads.


2

This kind of calculation is in general handled with using the joint distribution of the two variables $X$ and $Y$. You state that $X$ and $Y$ are i.i.d. $U(0, 1)$'s, and this uniquely specifies their joint distribution. Since they are independent, the joint density function factors: $$f_{X, Y}(x, y) = f_X(x) f_Y(y) = 1 \times 1 = 1$$ Given this, you can ...


1

The equality $Pr(T_1\leq t_1)=Pr(T_1\leq t_1\mid T_2> t_2)$ for every $t_1,t_2$ is indeed equivalent to independence of $T_1$ and $T_2$. And the initial equality does not need any independence assumptions. It is an identity. First, the events $T_1>t_1$ and $T_1\leq t_1$ are the opposite, so $$ Pr(T_1>t_1, T_2> t_2)+Pr(T_1\leq t_1, T_2> t_2) =...


1

Exponentials of numbers this large are beyond the edge of numerical stability. This is the trade-off of working with finite-precision arithmetic generally, not softmax specifically. The model you're using is a little strange because it treats counts (0,1,2,...) as if they are logits (any real). Softmax assumes that the inputs are on the logit scale. For ...


2

Conjugate priors may not be the best model choice if they are chosen for convenience (i.e., ease of deriving the posterior). Your priority -- ideally -- should be to select a prior that best describes your belief. For example, suppose you have a Poisson likelihood and have a prior belief that the rate parameter is uniformly likely to be (1,7), then you ...


2

Because the counts are very large. You actually calculate the following: $$\frac{e^{302}}{e^{302}+e^{51}+e^{247}}$$ which is of course not $1$, but extremely close to $1$ and Excel just rounds it. Note that the other numbers are also extremely small.


0

Firstly, your probability for A should be 13/72 because - hopefully it seems intuitively wrong to get more than half of the results be three, as implied by 13/24... Secondly, the probability for the union means either occur, not both. Thirdly, you should be calculating P(B|A). You condition on the three because this is what you observe. The probability for ...


1

Bayesian networks need to be acyclic. You don't put an arrow on every related node. Because, it is always the case that if $X$ is related to $Y$, $Y$ is related to $X$. Otherwise it'd be an undirected graph, or a directed graph with every relation doubled. Conceptually, causal relations make more sense when building the graph. And, the builder of the graph ...


0

Just an intuitive addition: when two RVs are iid, symmetry rules govern. For example, there is no reason that one RV is greater than the other with higher probability, i.e. $P(X>Y)=P(Y>X)$. Or, it's not expected that their expectations in the same form to differ. That means you can always switch variables, e.g. $E[X^2/Y]=E[Y^2/X]$.


2

If $X$ is independent of $Y$ and $X \sim Y$ then $X$ and $Y$ are exchangeable. That means that for any measurable function $f$, one has $f(X,Y) \sim f(Y,X)$, and consequently $E\bigl[f(X,Y)\bigr] = E\bigl[f(Y,X)\bigr]$ when this expectation exists. Apply this result to $f(x,y) = \frac{x}{x+y}$.


2

since $X, Y$ are iid then for joint distribution we have: $$\forall_{(x,y) \in \Omega} f_{XY}(x, y) = f_{XY}(y,x)$$ then: $$E[\frac{Y}{X+Y}] = \iint_{\Omega} \frac{y}{x+y}f_{XY}(x,y)dxdy=\iint_{\Omega} \frac{y}{x+y}f_{XY}(\underline{y,x})dxdy=^{Fubini Theorem}\iint_{\Omega} \frac{y}{x+y}f_{XY}(y,x)\underline{dydx}=^{x<->y}\iint_{\Omega} \frac{x}{x+y}...


1

It exists as long as the joint for all variables exist, and you can calculate it as follows: $$P(B,C)=\sum_{a\in\mathcal{A}}P(A,B,C)=P(B) \sum_{a\in\mathcal{A}} P(C|A,B)P(A)$$


2

I assume $X_{i}$ are all non-negative. (There is a way to treat general $X_{i}$, but then we need to delve into how we define expectations and such). This can be shown using the second Borel-Catelli lemma. The lemma states that given $E_{1},E_{2},\ldots$ independent events such that $\sum_{n=1}^{\infty} P(E_{n}) = \infty$, then $P(\limsup_{n} E_{n}) = 1$. ...


1

$p_\theta(\bf x)$ is the constructed likelihood function(intractable model density) and not marginal likelihood. The chain of partial derivatives indicates that the auxiliary function(CDF) used is differentiable in all dimensions. "Learning in Implicit Generative Models" (Mohamed and Lakshminarayanan, 2016) is a good reference concerning the learning and ...


0

Independence does not imply conditional independence: The simplest counter-example here is to take $C = \bar{A} \cup \bar{B}$ (i.e., the event $C$ occurs so long as $A$ and $B$ don't both occur). We then have the trivial result: $$\begin{equation} \begin{aligned} \mathbb{P}(A \cap B | C) &= \mathbb{P}(A \cap B | \bar{A} \cup \bar{B}) \\[6pt] &= \...


3

Since $e^{ t x}$ is a convex function, by Jensen's inequality, we have $$e^{t X} \le \frac{\sigma+X}{2\sigma}e^{-t\sigma}+ \frac{\sigma-X}{2\sigma}e^{t\sigma}$$ Taking expectation of both sides of above inequality we get $$E[ e^{t X}] \le \frac{1}{2}e^{-t\sigma} + \frac{1}{2}e^{t\sigma}$$ where we used the zero-mean assumption on $X$. It ...


2

1) $Y=F(X)$ is a transformation applied on $X$, just like $Y=X^2$ or $Y=\sqrt{X}$. So, $F(X)$ is not the PDF of $Y$, it's a transformation applied on $X$. The author chooses this transformation function to be the CDF of $X$ (for some reason of course, but that's not the question here). So, that is not the PDF of $Y$ (if it is, where is $y$ in it?). We just ...


3

Every moment/standardised moment is expressed in terms of some kind of expected value. For example, you've expressed the variance using expected value. But you couldn't do it using only the expected value of $X$, i.e. $E[X]$, since you cannot express $E[X^2]$ using $E[X]$ only. Similarly, the skewness can be expressed in terms of expectations as follows: $$E\...


4

This is the set $C_{m,n}$ of matrices that have at most one nonzero entry in each column. Let us rephrase the issue in the question as a property of an $m\times n$ matrix $A:$ $\mathcal{P}(A):$ for all independent random variables $x=(x_1,\ldots,x_n)$ defined on a given space $\Omega,$ $z=Ax$ is independent. My assertion requires two demonstrations: ...


3

For orthogonal columns, in general it's not. Let $$A=\begin{bmatrix}1 &-1\\1&1\end{bmatrix}$$ This results in $z_1=x_1-x_2,z_2=x_1+x_2$. These two aren't independent in general. For example, assume $x_i$ are Bernoulli RVs, then if $z_2=2$, $z_1$ is definitely $0$, which means there is dependency. For the linear independence, see @David's counter-...


0

Let's say your mixture components have probabilities $p_1,p_2$ (add up to $1$). A basic algorithm would probably choosing a uniform random variable between $u\in[0,1]$ and compare with $p$, if it's smaller sample from $\mathcal{N_1}$, else sample from $\mathcal{N_2}$. For $K$ components with probabilities $p_1,\dots,p_K$, do the same: If $u<p_1$, sample ...


0

The question has no unique answer. In particular, any putative formula based solely on the $p_{ij}$ cannot be generally correct. There are two reasons. The first is that when there is a positive chance of ties among any of the values, the answer will obviously depend on how frequently ties occur. Let's deal with that by assuming there is no chance of ties ...


3

Let $X$ be the number of tosses. Let $E[X]=M$ be the expected value of it. $X$ is $1$ with $0.7$ probability, and $2$ with $0.3\times 0.3=0.09$ probability. For all other cases, i.e. with $0.7\times 0.3 = 0.21$ probability, the whole process is repeated. So, we pay the two toss price and still expect $M$ tosses for the play to finish, which gives us the ...


2

It's not analytically available. You can use numerical approaches/software. What you want to find is actually inverse CDF of quantile function in other words. In Matlab you can use gaminv, in R, you can use qgamma, or in python you can use ppf in scipy.stats. It's not easy to find a table but here is one with unit scale, with varying shape parameters (upto ...


1

Let $H_1, H_2, H_3$ be the events we need to be independent. Any subcollection of events means that we can take any of these events. To be independent, they should satisfy four conditions: $$ k=2,\; i_1=1,\;i_2=2: \quad \mathbb P(H_1\cap H_2)=\mathbb P(H_1)\mathbb P(H_2). $$ $$ k=2,\; i_1=1,\;i_2=3: \quad \mathbb P(H_1\cap H_3)=\mathbb P(H_1)\mathbb P(H_3). ...


2

Begin with the hint given. Take $x=0.3$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.3\theta}{2}$? Take $x=0.8$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.8\theta}{2}$? Take $x=-0.5$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1-0.5\theta}{2}$? Take $x=-0.9$. Can you find the ...


1

The first step of finding the MLE is to write out the likelihood function (or log-likelihood function). In the present problem, with only a single observed data point, the latter is given by: $$\ell_x(\theta) = \ln(1 + x \theta) + \text{const.} \quad \quad \quad \text{for } -1 \leqslant \theta \leqslant 1.$$ This is the function you need to maximise to ...


0

Any two sample space can be combined by the operation of direct sum. That's generally denoted by the symbol $\oplus$, although the simple addition symbol is sometimes used when it's considered obvious that the direct sum is intended. However, in this case, it appears that regular addition is meant. The coin flips are being represented by the integers {0, 1}...


8

It's a good question, interpreted in the following way: the random variable $X$ determined by the coin experiment has a sample space of $\Omega_1=\{\text{Heads},\ \text{Tails}\}$ while the random variable $Y$ determined by the roll of a die has a sample space $\Omega_2$ consisting of the six possible stable orientations of the die on the table. These ...


0

Remember that X and Y will be the outcome of the roll of a dice/ flip of a coin. So if I understand your question correctly, you will ad the value you get from your first die roll (say 2) to the value of the coin toss (say 0). Resulting in a Z of 2 for example.


4

Surely they can have different sample spaces (or supports) but you can add them and generate another RV. If you assume independence between the two, the resulting mass function can be calculated by simply convolving the two. But, a more primitive solution is to list all possible scenarios: $$\begin{align}&P(Z=1)=P(Y=0)P(X=1)=1/12\\&P(Z=2)=P(Y=0)P(X=...


0

In order for it to be a proper probability distribution, $\pi$ should itself be a probability distribution, such that $$\forall k \quad \pi_k \geqslant 0$$ and $$\sum_{k=1}^K \pi_k= 1$$ Hence $$\int f(x) dx = \sum_{k=1}^{K}\pi_k \int N(x|\mu_k, \sigma_{k}^2) dx = \sum_{k=1}^{K}\pi_k \cdot1 = 1$$ and non-negative everywhere due to non-negativity of $\...


2

Here, it means the normal PDF: $$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$ The $\mu,\sigma^2$ in given side means that you can treat them as known quantities.


3

$N(x|\mu, \sigma)$ combines the two notations: $x \sim N(\mu, \sigma)$ and $p(x| \mu, \sigma)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma$.


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