Transformation of normal distribution

Question

Suppose $X$~$N(0,1)$ and we know $\Bbb E[X^{2n}]$. Now let $Y$~$N(0,\sigma^2)$.

I have received knowledge from someone that, given this information, we know that $$\Bbb E[Y^{2n}]=\sigma^2\Bbb E[X^{2n}]$$ Is this accurate? If so, how do we know this?

Answer 1

From the definition:

$$\mathbb{E}[Y^{2n}] = \int x^{2n}\frac{e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi \sigma^2}} dx $$ Let $t = x/\sigma$, so we can rewrite above as: $$\sigma^{2n} \int t^{2n} \frac{e^{-t^2/2}}{\sqrt{2\pi\sigma^2}}\sigma dt = \sigma^{2n} \int t^{2n}\frac{e^{-t^2/2}}{\sqrt{2\pi}} = \sigma^{2n} \mathbb{E}[X^{2n}]$$