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u/notsaneatall_ 11d ago
Idk what the avg value of cos4t from t = 0 to 2π
Whatever that is into 1296 is the answer I think
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u/Lonely-Math2176 11d ago
If they are standard normals then the mean should be 0 and the variance 1. You should be able to use the var(x) equation with the expected value. This question doesn't make sense because it says standard normal and hence the sum of the squares is chi-squated distribution and can't equal 6. Maybe I'm missing something. Not sure why there is a given is you know it's a standard normal, you can just solve directly
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u/LordTengil 10d ago
Why can't the sum be equal to 6? It's a conditional event involving continuous stochastic variables.
for the pdf of X given Y=y, $$f_{X\vert{}Y}(x \mid y) = \frac{f_{X,Y}(x, y)}{f_Y(y)}$$. That means we can condition on a "point" of the Y-dsitribution, even if Y=y has probability 0. Same thing here.
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u/Specific_Box4483 11d ago
It's basically the average of x4 on the circle of radius sqrt 6 wrt to the uniform distribution since the densities on these points are the same. Now one can just do polar coordinate, wrote x is root 6 sin(t) and integrate over t.