13

If you use the convention that $(\boldsymbol{x} - \boldsymbol{\mu})$ is a column vector, i.e. $(\boldsymbol{x} - \boldsymbol{\mu}) = \begin{bmatrix} x_{1} - \mu_1\\ x_{2} - \mu_2\\ \vdots \\ x_{m}- \mu_m \end{bmatrix}$, then $(\boldsymbol{x} - \boldsymbol{\mu})^T$ is a row vector, i.e $(\boldsymbol{x} - \...


8

Any function of the data is called an estimator. There is no such thing as "THE" estimator of a quantity. Various estimators can have different properties. You have shown (correctly) that your estimator $\tfrac{1}{n}\sum x_i^2$ is unbiased for $1+\mu^2$. You could consider other estimators and they may have different properties (e.g. smaller or ...


6

When you multiply matrices, the adjacent dimensions need to match, so you can multiply the (n, k) matrix by (k, m) matrix, or (m, k) by (k, n), but not any other way around. Where you would see the transpose symbol it depends on if the data is stored row-wise or column-wise. If you take something like a dot product of row vectors, you would transpose the ...


6

Hint: The variance of $X$ can be written as $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$.


5

Whether the distribution is skewed or not, the usual sample mean is always an unbiased estimator for the population mean.$^{\dagger}$ $$ \mathbb E\bigg[\bar X\bigg] = \mathbb E\Bigg[\frac{1}{n}\sum_{i = 1}^nX_i\Bigg]\\ = \frac{1}{n}\sum_{i = 1}^n \bigg( \mathbb E[X_i]\bigg) = \frac{1}{n} \sum_{i = 1}^n \mu = \frac{1}{n}n\mu = \mu $$ The usual sample mean is ...


5

You can use the properties of a non-central chi squared distribution to construct a non-biased estimator from the sum $$S_1 = \sum_{i=1}^n x_i^2$$ This sum $S_1$ has the mean $n+n\mu^2$. So $S/n$ will have the mean $1+\mu^2$, and is indeed an unbiased estimator. A more efficient estimator (an estimator with lower variance) is to be made with $S_2 = \left( \...


4

It is not clear what exactly the OP is asking. Consider, for example, the density of the vector random variable $(X_1, X_2, \ldots, X_n)$ whose component $X_i$ are $n$ independent standard normal random variables. The mode of this density (the location of the global maximum of the value of the density) is the origin. The expected value of the vector random ...


3

Notation is a tricky thing to do well and consistently. It is pretty rare in a book on any difficult stats topic that the notation is going to be perfect and consistent for the entire book. There is always a tradeoff between being precise (at the expensive of things looking complicated) and concise (at the expensive of not being precise). For expectation $E[...


3

In 1933 Bartlett described the Wishart distribution in terms of the distribution of the factors after Cholesky decomposition. I can not read the original source (On the theory of statistical regression) but this 'Bartlett decomposition' has been described elsewhere many times. See for instance the Wikipedia article on the Wishart distribution, which ...


3

For any column vector $x$ (eg $x \in \mathbb R^{n \times 1}$) $x^Tx$ is (a 1x1 matrix and thus 'is isomorphic to' (*)) a scalar. $xx^T$ is a matrix. (and if it's 1x1, then it could be treated as a scalar similarly.) (*) in your case 'is isomorphic to' just means 'can be treated as'


2

We have $$\mathbb E[h(X)]=\int_{-\infty}^\infty h(x) \frac{1}{\sqrt{2\pi}c^{1/2}}\exp\{-x^2/2c\}\,\text dx\\=\sum_{i=0}^n \int_{a_i}^{a_{i+1}} h(x) \frac{1}{\sqrt{2\pi}c^{1/2}}\exp\{-x^2/2c\}\,\text dx$$ by additivity of the integrals over disjoint intervals. And $$\int_{a_i}^{a_{i+1}} x h(x) \frac{1}{\sqrt{2\pi}c^{1/2}}\exp\{-x^2/2c\}\,\text dx=\int_{a_i}^{...


2

I can replicate your result by computing the mean as $$E[Y_{n}] = \sum_{y=1}^N P(Y_n \geq y) = \sum_{y=1}^N 1-\left( \frac{y-1}{N}\right)^n = N - \frac{1}{N^n} \sum_{y=0}^{N-1} y^n = N - \frac{H^{(-n)}_{N-1}}{N^n}$$ Then, if you approximate the generalized harmonic number with an integral $$\sum_{y=0}^{N-1}y^n \approx \int_0^{N-1}y^n dy = \frac{1}{n+1}(N-1)^...


1

Keep these in mind and you will soon find the answer: You know that the maximum number of draws is 10. If $M = 4$ means that you find ball $k$ on the 4th draw, then the expected number of draws is $\sum_{m=1}^{10} m \times P(M=m)$ $P(M=m)$ is the probability that you do not find ball $k$ on draws $1, 2, \dots, m-1$ and do find it on draw $m$.


1

Exploring Trying out a few values we see that the values get higher for larger $n$. (we have drawn a red line based on the bound computed below) f = function(n,p) { x = 0:n px = dbinom(x,n,p) x[1] = 1 return(sum(px*n*p/x)) } f = Vectorize(f) p = seq(0,1,0.001) plot(-1,-1, xlim = c(0,1), ylim = c(0,2), xlab = "p", ylab = "E[np/x]&...


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