# How to Find Probability Sum of Squares of Std Normal is greater than Sum of Squares of Non-Standard Normal with Mean 0

I'm looking for the probability that the sum of squares of a standard normal is less than the sum of squares of a non-standard normal with mean 0 and fixed std-dev.

Lets say $$X_i \sim \mathcal{N}(0,1)$$ and $$Y_i \sim \mathcal{N}(0,\sigma^2)$$ and there are 100 of each.

$$A=\sum_{i=1}^{100}(X_i)^2$$ which to my understanding becomes Chi-squared with 100 degrees of freedom.

But how about for $$B=\sum_{i=1}^{100}(Y_i)^2$$?

Is it as simple as just saying $$\sum_{i=1}^{100}(Y_i)^2 = \sum_{i=1}^{100}(\sqrt{1.5}(Z))^2$$ ~ $$1.5*{\chi}^2_{100}$$?

If we wanted $$P(B > A)$$, this would then be $$P(\frac{1.5{\chi}^2_{100}}{{\chi}^2_{100}} >1) = P(F_{100,100}>\frac{1}{1.5})$$.

Is this the correct solution or would B be some kind of gamma distribution and then I would need to scale it down to Chi-squared?

Assuming that $$\sigma$$ is known, you have $$\sigma^{-1}Y_i \sim \mathcal{N}(0,1)$$, so $$\sigma^{-2}\sum_{i=1}^{100} Y_i^2 \sim \chi^2_{100}$$. You then have $$\Pr(B>A)=\Pr\left(\frac{\sigma^{-2}B}{A}>\sigma^{-2}\right)=\Pr\left(F_{100,100}>\sigma^{-2}\right)\,.$$