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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 $EXY = (EX)(EY).$


3

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 $X=0$, ...


1

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


2

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 ...


14

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 ...


1

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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