4
The inequality should follow from the Law of Total Variance. Also, I'm assuming $\alpha \in [0,1]$, as otherwise the construction doesn't make sense.
We need a slightly more formal definition of your compound variable. Let $Z$ be an independent, binary variable with $P(Z=1)=\alpha$, $P(Z=0)=1-\alpha$, and thus we can define $R_2 := Z P + (1-Z)Q$ (I assume ...
3
$$E[(X-a)^3] =E[X^3 - 3X^2a+3Xa^2-a^3] = E[X^3] - 3aE[X^2]+3a^2E[X]-a^3$$
$$E[X^2]=\sigma_X^2+\mu_X^2$$
$$E[(X-a)^3] = E[X^3] - 3a(\sigma_X^2+\mu_X^2)+3a^2\mu_X-a^3$$
Only with $E[X^3]$ you can retrieve an exact result.
Without further information on $X$ it is not possible to say anything else on $E[X^3]$.
3
Yes, it is basically TRUE.
Mean is a first moment, the variance a second moment. Assuming moment estimates, the variance of an empirical mean is a second moment. The variance of an empirical variance is a fourth moment. So to ascertain the uncertainty in a variance estimate, you take fourth powers of your data! That will end to magnify errors quite a bit ...
...
3
For simplicity, assume $m=0$ and odd $n=2r+1$. The pdf of $Y_n$ is
$$f_n(y)={n\choose r} f(y) F(y)^r(1-F(y))^r$$
and we are interested in whether $\int y^2f_n(y)\,dy$ exists.
We know $F(y)\to 0$ as $y\to-\infty$ and $1-F(y)\to 0$ as $y\to\infty$.
If $F$ approaches 0/1 at some polynomial rate, then some power $r$ will approach faster than $y^{-2}$, ...
3
If you know the sample size (as opposed to just knowing that the sample size is between 50 and 200), then maximum likelihood might work in this case. Using Mathematica:
(* Joint distribution of largest 4 order statistics from a sample size of 100 *)
dist = OrderDistribution[{NormalDistribution[0, \[Sigma]], 100}, {97, 98, 99, 100}];
(* Log of the ...
2
The short answer is yes, in the case of fitting a straight line to a parabola. At the points where the expected fitted line intersects the true parabola, there is zero bias. The bias gets greater as you go farther away from the center of the $x$ data. (Just draw a straight line over a parabola; the vertical difference between the two is the bias.) In ...
2
As JohnK noted, the optimization problem that defines the ridge regression estimator can be defined without making any distributional assumptions on the noise; in particular the solution $\hat{\beta}_r = (X^\top X + \lambda I)^{-1} X^\top y$ remains the same.
I'll use the same notation $K = (X^\top X + \lambda)^{-1} X^\top X$ and $\hat{\beta} = (X^\top X)^{-...
2
To be clear, the solution to the ridge problem
$$ \min_{\beta}\left\{\frac 1 n \sum_{i=1}^n (y_i -x_i^T \beta)^2 + \lambda \|\beta\|_2^2\right\}$$
does not depend on the distribution of the error. Use, e.g., standard calculus to see this. This means that you do not have to modify the estimator if the error is not identically distributed. Sure, its ...
2
I can't follow your working after $\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\right)^2}{f(x;\theta)}d x$, note that this expression is nonnegative as the function being integrated is nonnegative.
Let $u = 1 + \theta \log x, \frac{du}{dx}=\frac{\theta}x$,
\begin{align}
&\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\...
1
The projection of $x$ onto the span of $v$ is a vector $\frac{x^\top v}{v^\top v} v$, but this vector has length $\frac{x^\top v}{v^\top v} \|v\| = \frac{x^\top v}{\sqrt{v^\top v}}$, which is what you should be using in your definition of variance. This will lead to $\frac{v^\top S v}{v^\top v}$ as the expression for variance (notice that replacing $v$ by $...
1
The equalities are incorrect, you should have $\sum Y_i^2 - n\bar{Y}^2$ on the RHS
First focus on the sum:
$$
\begin{aligned}
\sum_{i=1}^{n} (Y_i^2 - \bar{Y}^2) &= \sum_{i=1}^{n} [Y_i^2 - 2Y_i \bar{Y} + (\bar{Y})^2 ]\\
&= \sum_{i=1}^{n} Y_i^2 - 2 \bar{Y} \sum_{i=1}^{n} Y_i + (\bar{Y})^2 \sum_{i=1}^{n}1 \\
&= \sum_{i=1}^{n} Y_i^2 - 2n(\bar{Y})...
1
Since $\mathbb{E}[X]^2$ is non-negative, then the statement:
$$
\text{Var}(X) \leq \mathbb{E}[X^2]
$$
Is true in general. I am not sure what you mean by "dependent variables" here since there is only one random variable $X$.
1
The Maths is correct; the interpretation of what is the baseline needs a bit adjusting to get back to the EM result.
When comparing two arbitrary proportion coming from equal sized population, indeed the base-line proportion is $\bar{p}$ is $\frac{p_1 + p_2}{2}$. When though making an A/B test the base-line proportion is $p_1$. As such if we use $\bar{p}=p_1$...
1
Not a complete answer, but I hope it steers you towards it.
I'll show how it's possible to keep the matrices after the first derivative:
$$
\frac{\partial}{\partial\rho}\ln f_{\rho}(\boldsymbol x)=
-\frac12\frac{\partial}{\partial\rho}\ln|\Sigma| - \frac12 \frac{\partial}{\partial\rho}\left(\boldsymbol x'\Sigma^{-1}\boldsymbol x\right)
=
-\frac12\text{tr}((\...
1
When you classify $N$ samples and have a 50% chance of being correct each time, your TP+TN is a binomially distributed random variable $X$ with parameters $n=N$ and $p=\frac{1}{2}$. Wikipedia tells us that
$$ \text{Var}(X)=np(1-p)=\frac{N}{4}.$$
So
$$ \text{Var}(\text{Acc}) = \text{Var}\bigg(\frac{X}{N}\bigg)=\frac{1}{N^2}\text{Var}(X) =\frac{1}{4N}.$$
Also, ...
1
You are missing a lot of context here (presumably taken from linear regression) so I will focus solely on the asymptotic distribution:$^\dagger$
$$\sqrt N (\sigma^2 - \hat{\sigma}^2) \overset{\text{Dist}}{\rightarrow} \text{N}(0, 2 \sigma^4). \quad \quad$$
The left-hand-side here is a scaled version of the estimation error, which is affected by the size $N$. ...
1
Comment: This the germ of an idea, not yet a solution. [I am assuming the mean $\mu$ is also unknown.]
Suppose $n = 100$ observations from a standard normal distribution.
You might find the distance $D$ between order statistics 97 and 100.
Then $1/D$ should estimate $1.$ Several runs give various answers. In R,
set.seed(123)
diff(sort(rnorm(100))[c(97,100)])
...
1
There is no non-trivial upper bound that doesn't depend on a lot of detail about the matrix.
A conceptually simple version of the variance is
$$\mathrm{var}[\hat\beta]_{jj}=\frac{\mathrm{var}[Y|X]}{n\mathrm{var}[X_j|\textrm{other Xs}]}$$
Even fixing the distribution of each $X$ separately and of $Y$, the denominator can be an arbitrarily small positive ...
1
You are given data for two samples, so I suppose you are to find sample variances:
$S = \frac{1}{n-1}\sum_{i=1}^n(X_i - \bar X)^2.$
Before conducting formal tests, it is customary to look at various numerical
summaries of the data: Here are some numerical summaries from R statistical
software (thanks for providing data in a convenient format):
smoke = c(69....
1
No, assuming typical regression notation is used, $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n \hat{e}_i^2$ is an estimate for residual variance (since $\bar e = 0$), $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \overline{X})^2$ is an estimate for feature/independent variable variance, and $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (Y_i - \overline{Y})^2$ ...
1
I will show by example three density function all with zero mean, variance one, but different third moments (skew). As example I will use the standard normal distribution, and two skew-normal distributions with $\alpha$ parameter $\pm 3$:
Since these three distributions have the same first two moments, but different third moment, the answer should be clear.
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