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Let $X_1,...,X_n$ be a random sample from X with a Beta distribution $B(\theta; 2)$, with $\theta > 0$, i.e. : \begin{align*} f(x,\theta) = \theta(\theta+1)x^{\theta-1}(1-x) \mathbb{I}_{0<x<1} \en …

Imagine that: $C = \text{Convicted of rape} \\ I = \text{Innocent of rape} \\ A = \text{Accused of rape}$ If I want to know the probability of being innocent having been accused, I apply Bayes' theorem … of being accused while innocent for a random person is far under the probability of being innocent while accused ? …

This question is rather semantic than statistical. In Robust Statistics, estimators of mean and variance of a distribution are often called respectively "estimators of location" and "estimators of sca …

X_1=x1,...,X_n=x_n,T=t)}{P(T=t)}$$ According to the teaching assistant, this is equal to : $$ \frac{P(X_1=x1)...P(X_n=t-\sum_{i=1}^{n-1}x_i)}{P(T=t)}$$ I understand that because $X_1,...X_n$ are iid, probability … In other words, the probability that : "n random variables equal n observed values and sum of this n random variables is equal to the sum of the n observed values" is equal to the probability that n random …

Suppose an exponential distribution : $$ f(x) = \theta e^{-\theta x} $$ The MLE for theta is the inverse of the sample mean : $$ \frac{1}{\bar{X}} $$ I want to find the asymptotic normality of this es …

In this video, Blitzstein says that some random variables have no pdf but do have a cdf. Also, in my course material, I studied that converging in mean was stronger than converging in cdf which itself …