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Given a standard normal distribution $N(0,1)$ why, if you wish to scale it, you need to multiply by the square root of the variance? Ie given a variance t, the new scaled distribution is $N(0, t)$ which is equivalent to $\sqrt{t}*N(0,1)$?

Isn't the pdf of the general normal $\frac{1}{\sigma}$*(standard normal pdf)? An intuitive explanation and proof would be quite helpful.

shoestringfries
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1 Answers1

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Generally if $X$ is a random variable with mean $μ$ and Variance $σ^2$, then for any scalar $t$ you can prove that $E[tX]=tE[X]$ and $Var(tX)=t^2Var(X)$. So, if $X\sim N(0,1)$ and you want to create a random variable $Y\sim N(0,t)$ (where $t$ denotes the variance and not the standard deviation), then $$t=Var(Y)=Var(kX)=k^2Var(X)=k^2\cdot1=k^2 \implies k=\pm \sqrt{t}$$

Jimmy R.
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  • so my original note of the general normal formula being $\frac{1}{\sigma}$*standard normal pdf is irrelevant? – shoestringfries Nov 15 '16 at 13:11
  • Perhaps you confuse it with the reverse transformation: If $X\sim N(0,σ^2)$ with $σ>0$, then $$Z=\frac{X}{σ}\sim N(0,1)$$ – Jimmy R. Nov 15 '16 at 13:15