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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?

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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 $$ \Pr(B>A)=\Pr\left(\frac{\sigma^{-2}B}{A}>\sigma^{-2}\right)=\Pr\left(F_{100,100}>\sigma^{-2}\right)\,. $$

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