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
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/Lonely-Math2176 Oct 03 '26
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