Search Results

Find the joint probability function for $Y_A$ and $Y_B.$ This is essentially Exercise 5.1a in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer. … These correspond to the following options: $$ \begin{array}{ccc} Y_A &Y_B &Y_C\\ \hline 2 &0 &0\\ 1 &1 &0\\ 1 &0 &1\\ 0 &2 &0\\ 0 &1 &1\\ 0 &0 &2 \end{array} $$ Each is equi-probable, hence the joint probability …

We have \begin{align*} f_X(x) &=e^{-x}\\ f_Y(y) &=e^{-y}\\ f(x,y) &=\underbrace{f_X(x)\cdot f_Y(y)}_{\text{independent}}=e^{-(x+y)}\\ P\big(Y>X^2\big) &=\int_0^\infty\int_{x^2}^\infty e^{-(x+y)}\,dy\, …

Here's the algorithm I would run: Find peaks in the histogram. The data here is $$[0,0,1,4,5,3,0,1,0,0,0,0,1,6],$$ so that the index into the array corresponds to the original data. That is, there ar …

Yes, you can, mostly. There is a theorem, Theorem 5.12 in Wackerly, Mendenhall, and Scheaffer, 5th Ed., p. 228, that goes like this: Let $Y_1,Y_2,\dots,Y_n$ and $X_1,X_2,\dots,X_m$ be random varia …

I agree with the other answers, but I would like to add that in the New Causal Revolution, particularly with the highest level of reasoning - the counterfactual - it is possible to reason about indivi …

The distribution functions for $X_1$ and $X_2$ we write out as \begin{align*} f_1(x_1)&=\frac{x_1^{r-1}e^{-x_1}}{\Gamma(r)}\\ f_2(x_2)&=\frac{x_2^{s-1}e^{-x_2}}{\Gamma(s)}, \end{align*} with moment-ge …

I'm not used to your terminology, so let me break one of my own rules and use my terminology. Suppose we have a gamma random variable with parameters $\alpha,\beta,$ such that the density function is …

$\newcommand{\Cov}{\operatorname{Cov}}$ The claim is false, if we make the assumption that $\sigma^2\not=0.$ "Uncorrelated" occurs if and only if the covariances are zero. We know that \begin{align*} …

I understand how this works for conjunctions like the first probability. But I feel like that's a warmup question. …

You can interpret the correlation as a measure of how well a straight line (linear relationship) fits the data. If you have a correlation of $\pm 1,$ then it is a perfect linear relationship. But what …

Step 1. Taking the cue from whuber, we begin with a non-negative RV $X$ (we say non-negative because eventually $X=\left\|\hat{f}-f\right\|_n^2\ge 0$), and assume that $P(X\ge t)\le e^{-t},$ for all $ …

Each candidate has probability $1/m$ of being voted for by any voter, a constant. … \end{align*} To finish up, we note that the $(m-1)$th order statistic is going to be very similar - we can approximate it as the same probability as what we just calculated. …

To compute your desired probability, you need to calculate $$4\int_{0.5}^1\left[\int_0^y\big(y-x^2\big)\,dx\right]dy=0.854167.$$ …

I think Adam Kells has it mostly right, but let me write up a few more comments on your last question (which Adam calls Q4): The notation $Y_x(u)$ is referring to the value of the effect $Y$ when the …

It's not quite as simple as that: there is a $z$ in the inner sum so that the two sums are nested, not independent. Also, you need to be very careful about the order of operations. You really need to …