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1

By definition: $$cov(X, Y) = E(X-EX)(Y-EY) = EXY - (EX)(EY).$$ If $X, Y$ are independent, then $cov(X, Y) = 0,$ so that $EXY = (EX)(EY).$

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No, this equation is not valid. To see this, consider a simple example in which random variable $X$ is either $0$ or $1$, each with probability 1/2. Random variable $Y$ has the same distribution, but is perfectly negatively correlated with $X$. Therefore, $X^2$ has the same distribution as $X$, and $E[X^2] =1/2$. However, $E[XY]=0$ since whenever $X=0$, ...

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A number "a" increased by "b" means "a" plus "b", "a+b".

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Long form of angryavian's argument: Let $Y=\frac1aX=g(X)$, with $X$ having pdf $f_X$, and $a>0$. There is a standard way to find the pdf of Y: for any integrable function $h$, $$E(h(Y))=\int_{-\infty}^\infty h(y) f_Y(y) \mathrm{d}y$$ But you have also $$E(h(Y))=E(h(g(X)))=\int_{-\infty}^\infty h(g(x)) f_X(x) \mathrm{d}x=\int_{-\infty}^\infty h(x/a) ... 0 If you want your modified random number Y to follow the distribution \propto e^{-a^2 y^2}, the calculation there shows that aY follows the distribution \propto e^{-x^2}, so what you can do is generate a random number X from \propto e^{-x^2}, and then X/a will follow the distribution \propto e^{-a^2 y^2}. 0 To generate a numeric random variable onehas to know its distribution. The simple case of a uniform random variable in [0,1] can be simulated as follows: Mark a point on the boundary of a circular spinning top with perimeter length 1 and let it spin. Once it lands, measure the clockwise distance from the mark to the landing point. For other continuous ... 0 Asserting p to be a primitive root of prime q p is a primitive root of q iff p^n \mod q is different for each n in the range [0;q-2]. When p and q are internally primary integers greater than 1, p^n \mod q will always describe a repeating series beginning with 1 for n=0. No other value in the series is 1 until the series start ... 2 Using n unique characters, there are exactly n^k strings of length k. Think of it this way: You have n choices for the first character, n for the second one... n for the k-th one. So the amount of options is$$\underbrace{n·n·n\ldots n}_{k\text{ times}} = n^k$$If you want to know the amount of strings with n unique characters with ... 0 I apologize for the naive response, but I do not understand: Why not just choose a modified form of the questioner's suggestion? Why can you not generate N numbers from the distribution, then just normalize by multiplying each by the ratio of (desired_sum/actual_sum_of_the_numbers)? I'm not a math guy, but why won't this answer work? The best thing is that ... 0 This answer is trivially no for natural numbers. Let c=1 and d=2. Now P(ab=1)=\frac14, P(ab=2)=\frac12, P(ab=3)=0, and P(ab=4)=\frac14. 1 Here's a simple example: Suppose X and Y are independent and identically distributed uniform random variables in [0,1]. Let Z = XY. Clearly, we must have 0 \le Z \le 1. Thus, if Z were also uniform, we would have for instance \Pr[Z > 1/2] = 1/2. But XY \le X and XY \le Y, hence XY > 1/2 only if X > 1/2 and Y > 1/2; ... 1 If you treat each of them as random variables, then the product is also a random variable. To see this note that if X_1, \ldots, X_n are random variables, then (X_1, \ldots, X_n) is a random vector on \Bbb R^n, hence applyi,g the continuous, and therefore measurable function$$(X_1,\ldots, X_n)\mapsto X_1\ldots X_n We get the product to be the ...

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NOTE: This analysis only considers positive real numbers $c,d$ and independent random variables $a,b$ both uniformly distributed in $[c,d]$. In this case $ab$ is NOT uniformly distributed in $[c^2,d^2]$. Note that $(a,b)$ will be uniformly distributed in $S=[c,d]\times[c,d]$ whereas $ab=k\in[c^2,d^2]$ is a hyperbola that passes through the point $(\sqrt ... 0 No. Say you are choosing between 1 and 4. If you take every possible result you get: 1*1 = 1 1*2 = 2; 1*3 = 3; 1*4 = 4; 2*2 = 4; 2*3 = 6; 2*4 = 8; 3*3 = 9; 3*4 = 12; The result 4 is more likely appear than any other number and if the input is truly random as sample size increases it will become uniform. 2 Suppose after some draws you've seen$k$different colors. Furthermore, suppose you haven't seen any new colors in the previous$t$draws. If there are more than$k$colors, the probability that you see a new color within$t$draws is at least$1-\left(\frac{k}{k+1}\right)^t$. By letting$t$(the number of draws since the last ($k^{\text{th}}$) new color ... 1 If you draw$m$marbles, the chance that you not have seen a particular color is$\left(\frac {N-1}N \right)^m$, so the chance you have seen it is$1-\left(\frac {N-1}N \right)^m$. If we make the incorrect assumption that the colors are independent, the chance that you have seen them all is$\left(1-\left(\frac {N-1}N \right)^m\right)^N\$ This will be very ...

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The card must be fair in both cases, or in no one, because its behavior does not depend on the observer's knowledge: Case 1: an engineer knows whether the card is fair, flips it and tells you the result, Case 2: you do not know whether the card is fair, and flip it. Therefore, do not assume that the card is fair in case 2. You need a proof that it is ...

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