Hot answers tagged statistics
5
If you can factor the joint density into a product that is a function of x times a function of y, then $X$ and $Y$ are independent and their marginal densities are a constant multiple of the two functions in the product. That is, if
$$f_{X, Y}(x, y) = g(x) h(y)$$
for some functions $g(x)$ and $h(y)$, then $X$ and $Y$ are independent and
$$f_X(x) = c g(x)$$
...
4
The density of the individual $X_i$ is nearly the density of a normal distribution, but it has $\log x$ where one would expect $x$ and that additional $\frac{1}{x}$-factor. That means that the $X_i$ are log-normally distributed, i.e. $\log X_i$ is normally distributed. See http://en.wikipedia.org/wiki/Log-normal_distribution.
By taking the logarithm of the ...
3
This is generally easier to solve by the completary situation, that is: What's the probability of NOT being accepted by any universtiy? Then the one you're looking for will be $1-x$. Now that's easier to calculate because you just have to use the product rule: probabilities of each of them multiplied, as it's only one way of not being accepted anywhere, ...
3
Consider $$\begin{align*}\int\lambda xe^{-\lambda x}\,\mathrm{d}x&=\lambda\int\underbrace{x}_u\,\underbrace{e^{-\lambda x}\,\mathrm{d}x}_{\mathrm{d}v}\\&=\lambda\left(-\frac1\lambda xe^{-\lambda x}+\frac1\lambda\int e^{-\lambda x}\,\mathrm{d}x\right)\\&=\lambda\left(-\frac1\lambda xe^{-\lambda x}-\frac1{\lambda^2}e^{-\lambda ...
3
We have:
$$E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$$
Lets use Integration by Parts to do the indefinite integral and then we'll the limits of integration.
We have:
$$\int y^2\lambda e^{-\lambda y} dy = \lambda \int y^2 e^{-\lambda y} dy$$
using IBP, we let:
$u = y^2 \rightarrow du = 2 y dy$
$v = e^{-\lambda y} dy \rightarrow = ...
3
Integrating by parts, take $u= y^2 \Rightarrow du= 2y \cdot dy$ and $dv=e^{-\lambda y} \Rightarrow v = \dfrac{e^{-\lambda x}}{-\lambda}.$
$$\begin{align}
\int y^2\lambda e^{-\lambda y} dy & = \lambda \int y^2 e^{-\lambda y} dy \\
& = \lambda \left [\frac{y^2e^{-\lambda y}}{-\lambda} - \int \frac{2y e^{-\lambda y}}{(-\lambda)} dy\right] \\
& = ...
2
The sample mean is given by:
$M=\dfrac{\sum_{i=1}^n x_i}{n}$
where $x$ is your individual observations and $n$ the number of observations.
Before you added the observation you have:
$5=\dfrac{\sum_{i=1}^7 x_i}{7}$
And after you add the observation you have:
$6=\dfrac{\sum_{i=1}^8 x_i}{8}$
Given these you can solve for the sums of $x_i$
These turn out ...
2
Integrating by parts using $u=x \Rightarrow du=dx$ and $dv= e^{-\lambda x} dx \Rightarrow v = \dfrac{e^{-\lambda x}}{-\lambda}$, we have
$$\begin{align}
\int x\lambda e^{-\lambda x} dx & = \lambda \int x e^{-\lambda x} dx \\
& = \lambda \left [\frac{xe^{-\lambda x}}{-\lambda} - \int \frac{e^{-\lambda x}}{(-\lambda)} dx\right] \\
& = \lambda ...
2
Well, for better understanding how to find the sufficient statistic one cat see this very nice lecture.
Anyway, you can start with the joint probability density function of $\vec{X}=(X_1,\dots,X_n)$. So it will be $$f_{\vec{X}}(x_1\ldots x_n)=\prod_{i=1}^n\frac{1}{2x_i\sqrt{2\pi}}e^{-\frac{(\ln x_i - \theta)^2}{8}}=\prod_{i=1}^n\frac{1}{2\sqrt{2\pi}}e^{-\ln ...
2
Your random variable $X$ has pdf $f(x; \theta)$:
The joint density of $\overset{\rightharpoonup }{X}=\left(X_1,\ldots ,X_i,\ldots ,X_n\right)$, a random sample of size $n$ on $X$, is given by:
$$f_* \left(\overset{\rightharpoonup }{x};\theta \right)=\prod _{i=1}^n f \left(x_i;\theta \right)$$
The mathStatica function Sufficient constructs the joint ...
2
The function
$$
\lambda:x\mapsto \frac{f(x)}{1-F(x)}
$$
is called the hazard function (at least in survival analysis) and it measures the instantaneous rate of change in the sense that it can be described as
$$
\lambda(x)=\lim_{\Delta x\downarrow 0}\frac{P(X\in [x,x+\Delta x]\mid X>x)}{\Delta x}
$$
where $X$ is a random variable with distribution function ...
2
If you have $N$ independent random variables with densities $f_1,\ldots,f_N$, then the joint density is simply $$
f(x_1,\ldots,x_N) = f_1(x_1)\cdot\ldots\cdot f_N(x_N)
$$
The join density of $N$ independent random variables with $X_i \sim \textrm{Bin}(m,p)$ is thus $$
f(x_1,\ldots,x_N) = \prod_{i=1}^N \underbrace{\binom{m}{x_i} ...
2
In the 18th century Abraham de Moivre considered this problem (his account of which you can find in his book The Doctrine of Chances: If you toss a fair coin $1800$ times, what is the probability that the number of heads in in a specified interval, for example at least $880$ but no more than $906$? In the course of solving that problem, he first introduced ...
1
So, random variables $X$ and $Y$ have joint pdf $f(x,y)$:
which appears thusly:
Let $Z=X-Y$. Then, the cdf of $Z$ is $P(Z<z)$ = $P(X-Y<z)$:
where Prob is a mathStatica function.
Simply differentiate to obtain the pdf ...
1
No sure what you mean by "sl". You want a number $c$ such that $\Pr(\chi^2_{19}<c)=0.05$. If the test statistic is less than that, then you reject $H_0$; otherwise you don't. (The software I'm using gives $c=10.11701$.)
You're OK, except that as your bottom line, instead of concluding that the variance has not been reduced, you should conclude that ...
1
To understand the concept intuitively, ask the question of why it is needed. And start from a simpler question, why is the variance computed as $V = \frac 1 {N-1} \sum (x_i-\bar x)^2$ with a $N-1$ as denominator.
The fact is that you are interested in the variance of the population, not of the one of sample. Now obviously the sample is less dispersed than ...
1
I don't know if there is a smart way to solve the problem, but the standard way is the following:
for $1\le j< k\le n$, the joint density function should be
$$
f_{X_{j},X_{_k}}=\frac{n!}{j! (k-j-1)!(n-k)!}x^{j-1}(y-x)^{k-1-j}(1-y)^{n-k}
$$
where $0<x<y<1$
now to compute $E[X_{j}X_{_k}]=\int_0^1\mathbb{d}y\int_0^y xy\cdot f(x,y)\,\mathbb{d}x$, ...
1
You basically need to find the expected cost of each scheme.
The cost of that scheme now will depend on the cost incurred for each item sampled, and the cost for each batch rejected (sum of both expected costs).
The cost of each item sample in turn depends on the expected number of items sampled under the scheme.
The cost of each rejected batch depends on ...
1
$\displaystyle \int x\lambda e^{-\lambda x} dx$
$\displaystyle \lambda\int x e^{-\lambda x} dx$
Integrating by parts we have
$\displaystyle \lambda \left(x\int e^{-\lambda x} dx-\int\dfrac{dx}{dx}\left(\int e^{-\lambda x} dx\right) dx\right)$
$\displaystyle \lambda \left(\frac{xe^{-\lambda x}}{-\lambda}-\int \frac{e^{-\lambda x}}{-\lambda} dx\right)$
...
1
Let $\epsilon_i \sim N(0,\sigma^2)$. Then, we have:
$$Y_i \sim N(\beta_0 + \beta_1 X_i,\sigma^2)$$
Further clarification:
The above uses the following facts:
(a) Expectation is a linear operator,
(b) Variance of a constant is $0$,
(c) Covariance of a random variable with a constant is $0$ and finally,
(d) A linear combination of normals is also a ...
1
It's not enough. Let $X_i$ be independent uniform $[0,1]$ random variables and $f_i(x) = e^{i(X_i-x)^2}$. Then both $\sup_{x\in(0,1)}E(f_n(x))$ and $\sup_{x\in(0,1)}Var(f_n(x))$ converge to zero, but $\sup_{x\in(0,1)} f_n(x) = 1$ almost surely for every $n$.
Billingsley is a classic on this sort of thing. It's got all the examples of things like this you ...
1
I created a function based on root.
May $f_{money}$ be a function that depicts amount of money an agent possesses,
so that $f_{money}\left(u\right) \in \mathbb{N}, u \in \mathbb{A}$ where $\mathbb{A}$ is the amount of all agents.
Furthermore money in the system $m = \sum_{i=0}^{n-1}f_{money}\left(u_i\right) = 2500$
Number of agents $n = 5$
mean of money ...
1
One can check one's work on such things with a computer algebra system. For this problem, you have a random variable $X$ ~ Rayleigh($b$) with pdf $f(x)$:
Then, for random samples of size $n$ drawn on $X$, the CRLB for all unbiased estimators of $b^2$ is:
$$\text{CRLB}=\frac{1}{n *\text{FisherInformation}\left[b ^2,f\right]}$$
where FisherInformation is ...
Only top voted, non community-wiki answers of a minimum length are eligible
