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3

Unbiasedness by itself doesn't amount to much. Suppose $X_1,\ldots,X_n \sim \text{i.i.d. Bernoulli} (p)$, i.e. $$X_i = \begin{cases} 1 & \text{with probability } p, \\ 0 & \text{with probability } 1-p. \end{cases}$$ Then $X_1$ is an unbiased estimator of $p$, but it's a lousy estimator. On the other hand $(X_1+\cdots+X_n)/n$ is a far better ...


2

We calculate the cdf, dealing mainly with the case $a\ge 1$. Draw the line $y=1$. The joint density function lives in the part of the first quadrant that is below the line $y=1$. Draw the line $x+y=a$. Note that the geometry is a little different if $0\lt a\le 1$ than if $a\gt 1$, so making two diagrams is helpful. To find the probability that $X+Y\le a$ ...


2

Let us temporarily ignore the mechanic that a dead creature cannot be targeted. Relate this problem to the question of "If I flip twelve fair coins, what is the probability that I get more heads than tails?" (heads corresponding to hitting the 30 health minion, tails corresponding to hitting the 6 health minion) We could approach directly using the ...


2

The way to evaluate the nested expectations and variances is to regard $y$ as fixed (deterministic) in the inner calculation, then regard it as random in the outer. So for example, if you roll $y$ 6-sided dice (that are fair and numbered from $1$ to $6$), and $X$ represents the sum of the values rolled, then $$\operatorname{E}[X \mid Y] = \operatorname{E}[...


2

It is not true in general that $\mathbb{E}[e^{itX}]=e^{it\mathbb{E}[X]}$. For instance, suppose that $X=1$ with probability $\frac{1}{2}$ and $X=-1$ with probability $\frac{1}{2}$. Then $\mathbb{E}[X]=0$, hence $e^{it\mathbb{E}[X]}=1$. On the other hand, $$ \mathbb{E}[e^{itX}]=\frac{1}{2}\Big(e^{it}+e^{-it}\Big)=\cos(t) $$


2

You could do $$\mathbb{E}[(X^TX)^{-1}X^T\varepsilon] = \mathbb{E}[\mathbb E[(X^TX)^{-1}X^T\varepsilon \mid X]] = \mathbb E[(X^TX)^{-1}X^T\mathbb E[\varepsilon \mid X]] = 0$$ we used the law of iterated expectation and the fact that $(X^TX)^{-1}X^T$ is measurable wrt to $\sigma(X)$ (so we can bring it outside the conditional expectation)


2

I will try to start from the simplest case possible and then build up to your situation, in order to hopefully develop some intuition for the notion of convolution. Convolution essentially generalizes the process of calculating the coefficients of the product of two polynomials. See for example here: Multiplying polynomial coefficients. This also comes up ...


1

For convenience, define the matrix $M = (I+XA)$. Now use the Frobenius (:) Inner Product to write the function, differential and gradient as $$\eqalign{ f &= \log(\det(M)) \cr &= {\rm tr}(\log(M)) \cr\cr df &= M^{-T}:dM \cr &= M^{-T}:dX\,A \cr &= M^{-T}A^T:dX \cr\cr \frac{\partial f}{\partial X} &= M^{-T}A^T \cr &= (...


1

Here's the easy way: People are passing by the store with a constant average rate of $100$ per that interval, and each one has an independent probability of entering the store of $0.15$.   Therefore people are entering the store at a constant average rate of $15$ per that interval. The hard way (what you were trying): $$\begin{align}\mathsf E(X) =&...


1

Your relationship is not correct unless $\bar{x} = 0$ or $n=1$. Try it with $n=3, x_i = i$. Here $\bar{x} =2$, $\sum x_i(1-x_i) = -8$, and $\sum(x_i-\bar{x})^2 = 2$. Your relation then reads $$ -4 - (-16) = 4 $$ In fact, the discrepancy is always $$2\left( \frac1{n}-1\right)\sum_{i=1}^nx_i$$


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First of all, it is worth to specify that you are talking about normal distribution. Otherwise, $S^2$ is not (necessarily) the MLE of $\text{var}(X)$. "if the MLE is supposed to reflect the best attempt..." There is no universally best method to derive estimators. ML maximization is only one possible and widely accepted method. However, its ...


1

I doubt there can be a definite answer unless we find the person who wrote it there and asked him or her. I am guessing that $G$ refers to Gaussian. Could $\mu+C(\sigma)$ be a confidence interval? and the person might want to know what is the image of this interval under exponential map?


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$Y=X^2+Z$ means $Z=Y-X^2$, so we let $z(x,y)=y-x^2$ After affirming that there is a bijection between $(X,Z)\leftrightarrow(X,Y)$ , (because why?), we then can directly apply the Jacobian change of variables transformation: $$\begin{align}f_{X,Y}(x, y) =&~ f_{X,Z}(x, z(x,y))~\big/\Big\lvert\dfrac{\partial\big(x,z(x,y)\big)}{\partial\big(x,y\big)}\Big\...


1

I'm going to follow the notation you used initially and use capital $X$ and capital $Y$ rather than $x$ and $y$ for the random variables. $Y=1,\,2,\,3,\,\text{ or } 4$ each with probability $\dfrac 1 4$. $\operatorname{E}(Y) = 2.5$ and $\operatorname{var}(Y) = 1.25$. $X$ is the sum of the outcomes when $Y$ dice are thrown. The outcome when one die is ...


1

So you want to know the probability of obtaining $r$ specific distinct numbers by unbiased selection of $n$ from $N$ numbers with replacement. The Stirling number of the second kind $\begin{Bmatrix}n\\r\end{Bmatrix}$ counts the ways to partition $n$ unlabelled items into $r$ non-empty unlabelled subsets. $$P=\dfrac{\begin{Bmatrix}n\\r\end{Bmatrix}r!}{n^N}$...


1

So we know that we sampled $n$ times, and we know the numbers that came up, but not their respective multiplicities. Arrange the numbers we got in order, and let their multiplicities be $x_1$ to $x_r$ respectively. Then $x_1+\cdots +x_r=n$. By Stars and Bars the number of possibilities is $\binom{n+r-1}{n}$. Note that these possibilities are not all ...


1

Imagine two urns and 12 balls. After you distribute the balls randomly between the urns you take all but 6 from the second urn and move them to the first one (or move none if the second urn has 6 or less balls in it). Your question is if the second urn 5 or less balls in it (then the minion would be still alive). However, moving the balls does not change ...


1

You have two entities (call them Thing1 and Thing2) with powers $A$ and $B$. You generate independent uniform random numbers $a$ on $(0, A)$ and $b$ on $(0,B)$. Thing1 wins if $a>b$ and Thing2 wins if $a< b$ (we don't worry about equality of $a$ and $b$ as this occurs with probability $0$, but if it does worry you draw again). The outcome is uniformly ...


1

You are using $x$ in two places, one for the range of the random numbers and one for the number picked. I will use $X$ and $Y$ for the numbers picked and $x$ and $y$ the ranges they come from. If you draw a rectangle that has its lower corner at the origin, extends $X$ units to the right and $Y$ units up, you are picking a random point inside the rectangle....


1

The relationship, for discrete variates, of the (discrete) probability function (sometimes called probability mass) and the analog of the CDF, is the same as for continuous variates, but replasing integration with summation (and differentiation with first differences. You have to be meticulous about whether the endpoint is included in the sum or not (it is ...


1

If the random variable $Y$ takes only integer values, then $$ \mathbb{P}(Y=k)=\mathbb{P}(Y\leq k)-\mathbb{P}(Y\leq k-1)$$ Note that for a discrete random variable, the usual terminology is probability mass function rather than probability density function.


1

Let $X_1,..,X_n$ i.i.d $\mathcal{N}(\mu, \sigma^2)$ each one,hence by definition $$ \sum_{i=1}^n\left(\frac{X_i - \mu}{\sigma} \right)^2 \sim \chi^2_n \, . $$ Now, in order to estimate the variance of the Normal distribution, you are using (variation of) $$ S^2 = \frac{1}{n}\sum_{i=1}^n(X_i - \bar{X})^2, $$ thus $$ \sigma^2 S^2 = \frac{\sigma^2}{n}\sum_{i=...


1

One may recall that $$ \bar{x}=\frac1N\cdot \sum_{n=1}^Nx_n $$ giving $$ \begin{align} N^2\cdot \frac1N\sum_{n=1}^N(x_n-\bar{x})^2&=N\sum_{n=1}^N(x_n^2-2x_n\cdot\bar{x}+\bar{x}^2) \\\\&=N\sum_{n=1}^Nx_n^2-2\:N\cdot\bar{x}\sum_{n=1}^Nx_n+N\cdot N\cdot \bar{x}^2 \\\\&=N\sum_{n=1}^Nx_n^2-2\:\left(\sum_{n=1}^Nx_n\right)^2+\left(\sum_{n=1}^Nx_n\right)^...


1

We first give a brief outline of the theory. Let $F_X(x)$ be the probability that the measurement is $\le x$. Let $F_S(x)$ be the probability that the measurement is $\le x$, given that it was taken with the first machine, and let $F_T(x)$ be the same thing for the second machine. Then $$F_X(x)=(0.5)F_S(x)+(0.5)F_T(x).$$ Differentiating we find that $$f_X(x)=...


1

In your problem, there are five independent experiments, each of which is the sum of two die rolls. This is different from ten dice rolls. For example, you would expect a mean of 7 from your experiment, and 3.5 from the single dice rolls. In excel, create two columns of five rows of random die rolls (=INT(RAND()*6)+1 in cells A1..B5), and then add the ...



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