PDF of sufficient statistic of a normal distribution $N(0, theta)$

First observe that $frac{X}{sqrt{theta}}sim N(0;1)$ and thus

$$ bbox[5px,border:2px solid black] { frac{X^2}{theta}sim chi_{(1)}^2=GammaBig(frac{1}{2};frac{1}{2}Big) qquad (1) } $$

Now it is evident that

$$ bbox[5px,border:2px solid black] { T=frac{1}{theta}sum_{i=1}^{n}X_i^2sim GammaBig(frac{n}{2};frac{1}{2}Big) qquad (2) } $$

Concluding, $Y=theta T$,

thus

$$ bbox[5px,border:2px solid black] { Ysim GammaBig(frac{n}{2};frac{1}{2theta}Big) qquad (3) } $$

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To prove (1) and (3) use the fundamental transformation theorem (change of variable)

To prove (2) use MGF's properties

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Some hints for the proofs:

For (1), use the change of variable

$Z=frac{X^2}{theta}$ then

$$F_Z(z)=mathbb{P}[Zleq z]=mathbb{P}[X^2leq ztheta]=mathbb{P}[-sqrt{ztheta}leq X leq sqrt{ztheta}]=F_X(sqrt{ztheta})-F_X(-sqrt{ztheta})$$

derivate it and get you first PDF:

$$f_Z(z)=frac{1}{sqrt{2pi}}z^{-frac{1}{2}}e^{-frac{z}{2}}$$

This can be rewritten in the following way:

$$f_Z(z)=frac{Big(frac{1}{2}Big)^{frac{1}{2}}}{GammaBig(frac{1}{2}Big)}z^{frac{1}{2}-1}e^{-frac{z}{2}}$$

...and the first step is done! $f(z)$ is evidently a $GammaBig(frac{1}{2};frac{1}{2}Big)$

For step (2) very easily multiply the n identical MGF's

For step (3) same procedure as (1): change of variable.