Why is it that $N(0, ct) = \sqrt c N(0,t)$? What does it mean when we take a constant out of a distribution?
2026-02-23 05:57:35.1771826255
Scaling Normal Distribution
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It means that $X \sim N(0,t)$, i.e. $X$ is a random variable with normal distribution with mean $0$ and variance $t$, if and only if $\sqrt{c} X \sim N(0,ct)$. Multiplication by $\sqrt{c}$ always multiplies the variance of a random variable by $c$ and the mean by $\sqrt{c}$. The normal distributions are rather special in that multiplication by a constant preserves the fact of being normal; this can be seen from considering the density.