# Thread: joint distribution of the sample mean and variance

1. ## joint distribution of the sample mean and variance

suppose that the random variables x1,x2, and x3 are i.i.d., and that each had a standard normal distribution. also suppose that
y1= .8x1 + .6x2
y2= [sqrt.(2)]*(.3x1-.4x2-.5x3)
y3= [sqrt.(2)]*(.3x1-.4x2-.5x3(
find the joint distribution of y1, y2, y3.
i feel like this should be simple but i have no idea how to do it. my book says the joint distribution is 1/detA * the joint distribution of the inverse of A * y. No idea how to start this problem any help would be great

2. see http://en.wikipedia.org/wiki/Multiva...l_distribution

the point is that, the y's will be a trivariate normal, so all you need to do is compute the
mean and variance of that vector. The mean will be zero, since the mean of the X vector is zero.
Next you need the variance/covariance matrix of the Y's

3. Sorry, what? I've gotten as far as the joint dist. of X is:
1/(2pi)^n/2 e^(.5 * sum of X^2)

4. I've been trying the internet for a while and it's not helping. I'm just completely new to this stuff

5. Thanks, how do I start to solve for variance?

6. $V(AX)=AV(X)A^t=AI_3A^t=AA^t$ since $V(X)=I_3$

A is a 3 by 3 matrix with first row (.8,.6,0)
You're just obtaining the 3 variances and the 3 covariances.

7. I got -.18 for the determinant, so now what? Sorry for so many questions

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### suppose that the random variables x1, x2, and x3 are i.i.d., and that each has the standard normal distribution. also, suppose that. y1 = 0.8x1 0.6x2, y2 = √ 2(0.3x1 − 0.4x2 − 0.5x3), y3 = √2(0.3x1 − 0.4x2 0.5x3). find the joint distribution

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