Hot answers tagged

13

First, you might want to look at Wikipedia on Irwin-Hall distribution. Unless $n$ is very small $A = \bar X = \frac{1}{n}\sum_{i=1}^{n} X_i,$ where $X_i$ are independently $\mathsf{Unif}(\theta-.5,\theta+.5)$ has $A \stackrel{aprx}{\sim}\mathsf{Norm}(\mu = \theta, \sigma = 1/\sqrt{12n}).$ [The approximation is quite good for $n \ge 10.$ In fact, in the ...


9

No, it's not uniform. Intuitively, you would expect that the uncertainty over $\bar X$ decreases as $n$ increases. Also the central limit theorem suggests, as $n$ increases, the distribution approaches normal distribution. Which means, you'll have a peak around $\theta$, and it's going to narrow down as $n\rightarrow\infty$. For a simple counter-example, if ...


6

You'll use law of iterated expectations: $$E[X]=E[E[X|Y]]=E[Y]=\frac{10}{10-2}=5/4$$


3

What you are told is only relevant when you are sampling from a finite population of size $N$, with simple random sampling without replacement. For most applications, there is no definite finite population, so what you are told is irrelevant. It is also irrelevant when you are samling with replacement. When relevant, there is a finite population ...


2

The mathematical definition of an unbiased estimator is: $E[u(X_1, X_2,\dots,X_n)]=\theta$. In English, this formula means that the expected value of a statistic, generally given as $u(X_1, X_2], \dots, X_n)$, equals the parameter (of the population) value. While a parameter has a single (probably unknown) value, a statistic has a distribution of values (...


2

It can be shown via variable addition and subtraction: $$\begin{align}\sum (y_i-\mu)^2&=\sum(y_i-\bar y +\bar y -\mu)^2\\&=\sum(y_i-\bar y)^2+\sum(\bar y-\mu)^2 + \sum2(y_i-\bar y)(\bar y - \mu)\\&=\sum(y_i-\bar y)^2+n(\bar y-\mu)^2 + 2(\bar y - \mu)\underbrace{\sum(y_i-\bar y)}_0\\&=\sum(y_i-\bar y)^2+n(\bar y-\mu)^2\end{align}$$


2

Usually, measurement errors are reported as some multiple of the standard deviation: often the SD itself, or twice the SD, or occasionally some other multiple. To cover all these cases, let's just call that multiple $\kappa.$ In this fashion, taking the measurement errors to be independent, we may conceive of the numbers $x_i$ in your dataset as being of ...


2

The Irwin-Hall distribution is the distribution of a sum of $n$ uniform random variables. Therefore, an analytic expression for the density of the mean of $n$ uniform random variables is $$\frac{1}{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} (x-k)_+^{n-1}$$ By shifting this expression, you get the density of yours.


1

Via law of iterated expectations, it's equal to $$\mathbb E[\max_{0<t<1} X(t)]=\mathbb E[\mathbb E[\max_{0<t<1}X(t)|A,B]]=\mathbb E[\max (A,B)]$$, which you can use the formula found in Equation (11) of this paper. Note that the answer involves the CDF function $\Phi(t)$ of standard normal RV.


1

Ideally, you need the original measures. For example, for your 7 +- 0.5, you should have something like. [6.5, 7, 7.5, 7 , 6.5, 7.5 etc] If you do not have them then try to "Unroll" them yourself in a similar way, for example take 1.000 random samples with mean 7 and variance 0.5. Then you will have one bigger sample, which can be analyzed, further. ...


Only top voted, non community-wiki answers of a minimum length are eligible