# Tag Info

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No, $P(A \cap B) \neq P(A)P(B)$ For example, if $A$ is the event "when rolling a dice, it comes out 6" and $B$ is the event "it comes out 5" you can see that if you know that B happened (ie a 5 has been rolled) then you know that A has not happened In this case $P(A|B) = 0 \neq P(A) P(B)$ If B does not give you any information about A, then we say that A ...

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Here is a piecewise formula for $f_z$. $$f_z(z) = \begin{cases} z+1 & -1\leq z\leq 0 \\ 1-z & 0\leq z\leq 1 \\ 0 & \text{else}. \end{cases}$$ Computing a formula for $F_z$ is just a matter of integrating. $$F_z(z) = \int_{-1}^z f_z(z) \, dz = \begin{cases} 0 & z\leq -1 \\ \frac{z^2}{2}+z+\frac{1}{2} & -1<z\leq 0 \\ ... 1 If P(X=Y)=1 then for any measurable A we have:$$P(X\in A\wedge X\neq Y)\leq P(X\neq Y)=0$$so that P(X\in A\wedge X\neq Y)=0 and consequently:$$P(X\in A)=P(X\in A\wedge X=Y)+P(X\in A\wedge X\neq Y)=P(X\in A\wedge X=Y)$$Likewise we also find:$$P(Y\in A)=P(Y\in A\wedge X=Y)$$This can be applied to find:$$P(X\in A)=P(X\in A\wedge X=Y)=P(Y\in ...

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The expectation of a random variable is just a number, whereas uniform convergence refers to a sequence of functions. The author probably meant almost surely, that is: $P(\lim_{n\to \infty}\bar X_n=E(X))=1$ which means that $\forall \epsilon>0, \;\exists k\in \mathbb{N}: |\bar X_n-E(X)|<\epsilon \; \forall n>k$ (except for some set of infinitely ...

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Since Y is a positive random variable, only outcomes with w=x+y>x=z are possible. Because all positive values are possibilities for Y and X, the pdf will be posive for all z=x>0 and all w=x+y>x=z. It follows that pdf is positive at {(z, w): z>0, w>z}. No need to compute the pdf to determine where pdf is positive in above reasoning. In fact, the above ...

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Counterxample: Let $X$ be distributed as a mixture of two variables $Y,Z$ with some known densities and mixing paramenter $\theta \in [0,1]$. Then $p_X(x_i) = \theta \, p_Y(x_i)+(1-\theta)\, p_Z(x_i) = \theta \,g(x_i)+h(x_i)$ and $$L(\theta)=\prod_{i=1}^n (\theta \,g(x_i)+h(x_i))$$ This is a polynomial of degree $n$ in $\theta$, so it can have several ...

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A direct method would rely on induction: Let $Y \sim \operatorname{NegBinomial}(r,p)$ with $$\Pr[Y = y] = \binom{r+y-1}{y} p^r (1-p)^y, \quad y = 0, 1, 2, \ldots,$$ so $Y$ counts the random number of "failures" before the $r^{\rm th}$ "success" is observed, where the probability of "success" is $p$. Let $X \sim \operatorname{Geometric}(p)$ with $$\Pr[X = ... 1 It's a double summation.$$\sum_{n=1}^N\sum_{n'=1}^N$$Of course the given notation is more convenient. 1 Expected value of a quotient is not necessarily the quotient of the expected values. 1 For the geometric version, you drew a picture, which will serve us well for the integral. I would rather call the random variables X and Y, but draw the y_1-axis where the x-axis usually is, and draw the y_2 axis where the y-axis usually is. We are integrating over a certain region, which is actually a trapezoid. Let us integrate first with ... 1 You don't need any calculus: it's a matter of simple algebra. If x>0 and y>0, then 0<x<x+y, and since x+y>0, we can divide by x+y to get 0<\frac{x}{x+y}<1. (And you don't mean the minimum and maximum of a bivariate "equation." You mean bounds on the values of a bivariate function.) 1 Following and improving on Julia's line of thought, here's the function to calculate the number of lists of length n at Kendall-Tau distance d from some permutation: public static int getCount(int n, int d) { if (d < 0) return 0; if (n < 1) return 0; if (d == 0) return 1; if (d == 1) return n-1; if (d > n * (n-1) / 2) return 0; ... 1 You seem to have an error:$$|X_n - 1| < \epsilon \quad \iff \quad 1-\epsilon < X_n < 1+\epsilon.$$1 The random variables \bar{X}-\bar{Y}\,,\,\, X_i-\bar{X} and Y_i-\bar{Y} are jointly (and marginally) normal, since they are linear combinations of independent, Normally distributed, random variables. Consequently, a well known property of Multivariate Gaussians guarantees that$$ \begin{eqnarray*} \mathbb{C}ov\left(\bar{X}-\bar{Y}\,,\, ...

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Here's one way of viewing it. We want to write $$\begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} = \hat\alpha\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} + \hat\beta \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} + \begin{bmatrix} \hat\varepsilon_1 \\ \vdots \\ \varepsilon_n \end{bmatrix}$$ and choose the values of $\hat\alpha$ and ...

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Notation: $\langle x,y \rangle$ is the inner product between $x$ and $y$. It is sometimes called the dot product, denoted by $x \cdot y$. In any case we have $\langle x,y \rangle = \sum_{i=1}^n x_i y_i$. You want an approximate solution to $Ax=b$. The trick is to choose it such that $Ax-b$ is orthogonal to the span of $A$. In other words you want $\langle ... 1 Answer is 1/9.for example you take a 5 digit number then there are 9*10*10*10*10 ways then for successive number to have same digit there are 1*10*10*10*10 ways.probability is 1/9 by dividing 1 A more expressive model comes at a cost of statistical power and risks overfitting. For example, if you have a 10,000,000 degree polynomial statistical regression model, you can fit almost any kooky set of data, as long as n<10,000,000. But so what? Observed fit is not equivalent to predictive power. Simple models, even those that are "wrong", can ... 1 First, I am assuming the project is about or heavily involves mathematics. To come up with a good project you should find a problem that needs to be addressed (problem statement), decide on how your project will address that problem (purpose statement), and be able to explain why what you are doing matters. Here are some slides that go over this process. ... 1 Let's get you started. Suppose without loss of generality that$\mu_1 \leq \mu_2$. Then we have that $$\mu = x_1\mu_1 + x_2\mu_2 \leq (x_1 + x_2)\mu_2 = \mu_2.$$ Each other subcomponent of the proof looks extremely similar to this line. 1 It is because you have$6$numbers as$2^x$in the range$[0,100]$. For the same range you have$50$even numbers. Since the probability of every number to happen is equal, you have$1/6$for powers and$1/50$for the even number. But I think your set must be restricted to$[0,100]$, else these probabilities will change. 1 Lets look at it like you need a total of$60$percentage points. Right now you have $$.05(60)+.05(53)+.3(47)+.1(66)=26.35\text{ percentage points}$$ So, you need to now achieve $$.05(\text{A3})+.05(\text{A4})+.3(\text{T2})+.10(\text{P2})$$ such that the sum of that expression is$33.65$. If you are looking for what single grade could you score on them ... 1 You can certainly say that someone in group A is$\frac {77}{52}\$ times more likely to like singing than someone in group B.

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