Recent content by das1

OK thank you I'll try this

The problem: Let the five numbers 2,3,5,9,10 come from the uniform distribution on [$\alpha$,$\beta$]. Find the method of moments estimates for $\alpha$ and $\beta$ . I am trying to wrap my head around the idea behind estimates of moments. From what I understand, the first moment is the mean...

Ah ok thank you. Guess there's no reason you'd ever need to complicate your life by doing that.

But if the density is 1 between 1 and 2, doesn't that mean that it should be 0 everywhere else? And that's not true--If I integrate with different limits, say between 1 and 3, I get $\frac{4}{3}$. If the density were 0 outside this interval, doesn't that mean I should still get 1 between 1 and 3?

Hi! Thank you you've been very helpful. Integrating between 1 and 2 gets the indefinite integral of $-\frac{2}{x}$. Between 2 and 1 this is (-1) - (-2) = 1. Which makes sense, but how do I take that info and figure out the probability that W < 1.5 ? Thanks again

Looking at another textbook problem, hope someone can let me know if I'm on the right track: Let $X_1, X_2, ... X_{25}$ be a random sample from some distribution and let $W = T(X_1, X_2, ... X_{25})$ be a statistic. Suppose the sampling distribution of W has a pdf given by $f(x) =...

Hi, thank you! That's what I suspected, I appreciate the affirmation

The problem: Researchers conduct a study to test the effectiveness of a drug preventing a disease. Of 20 patients in the study, 10 are randomly assigned to receive the drug and 10 to receive a placebo. After 1 year, suppose 5 patients in the control group contract the disease, while 2 patients...

OK, did it all out and got $$\frac{-var(X)}{2} = cov(X,Y)$$ then $$-var(X) = 2cov(X,Y)$$ then $$V[X+Y] = V[X] = V[Y]$$ I was thinking I needed a constant but not sure if that's possible here...is there a way I can actually make this a constant or is this as simple as it can get?

Hi thank you! So substituting that we have something like $$-0.5 = \frac{Cov(X,Y)}{SD[X]^2}$$ and $$V[X+Y] = 2V[X] + 2Cov[X+Y]$$ Wouldn't we still need to know more info, like what Cov(X,Y) is, before solving? And we can't get those without knowing what SD[X] or V[X] are (or SD[Y] or V[Y])...

Suppose X and Y have the same distribution, and Corr(X,Y) = -0.5. Find V[X+Y]. I know that Corr(X,Y)*SD[X]SD[Y] = Cov(X,Y) and also V[X+Y] = V[X] + V[Y] + 2Cov(X,Y) So there must be a missing link, maybe an identity, that I'm not realizing. I think the fact that the 2 variables have the same...

Can someone explain this to me? Thanks! The component in the ith row and jth column of a matrix can be labeled m(i,j). In this sense a matrix is a function of a pair of integers. For what set S is the set of 2 × 2 matrices the same as the set Rs ? Generalize to other size matrices.

I need help on this problem, anyone know how to do it? Suppose you have n independent observations from a uniform distribution over the interval [𝜃1, 𝜃2]. a. Find the maximum likelihood estimator for each of the endpoints θ1 and θ2. b. Based on your result in part (a), what would you expect...

OK so this is what I've done: I set up the integral from 0 to 2 of (x+y^2)dx dy = 1 I've never done an integral with 2 variables before but I plugged it into this calculator: Integral Calculator - Symbolab and it gave me 28/3 ? Set that = 1 and divide and get 3/28? That's my best guess so far...

Consider the following joint probability distribution function of (X , Y): a(x + y^2) {0<=x<=2, 0<=y<=2} 0 otherwise Calculate the value of the constant a that makes this a legitimate probability distribution. (Round your answer to four decimal places as appropriate.) And then, For the...