# Standardized residuals vs. regular residuals

I've got an easy question concerning residual analysis. So when I compute a QQ-Plot with standardized residuals $\widehat{d}$ on the y-axis and I observe normal distributed standardized residuals, why can I assume that the error term $u$ is normal distributed? I'd think that if $\widehat{d}$ looks normal distributed I just could assume that the standardized error term $d$ should be normal distributed.

So why can we assume that $u\sim N$ when we just observe that $\widehat{d}\approx N$. By the way can we assume it?

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could you please mention how you calculate $u$ and $\hat d$ –  suncoolsu Jul 12 '11 at 16:41
Hi. you can't calculate $u$ (except you're observing the whole population). So I'd say I calculate $\widehat{u}=y-\widehat{y}$ and $\widehat{d}=\widehat{u}/\widehat{\sigma^2}$. I guess I've got an idea of the answer, but I can not show it. What I know is, that we often assume that $u\sim N(0,\sigma^2)$, so when we divide $u$ by $\sigma^2$ I assume that we divide the first 2 moments of the distribution as well $u/\sigma^2 \sim N(0/\sigma^2,\sigma^2/\sigma^2)$ where $u/\sigma^2=d$ ist the std.error term which is (from above) $d\sim N(0,1)$. But I'm not sure if this makes sense (That's why I ask) –  MarkDollar Jul 12 '11 at 19:46
... Thus when $u\sim N(0,\sigma^2)$ we assume automatically that $d\sim N(0,1)$. Is this correct? And where can I find a proof that this is correct? –  MarkDollar Jul 13 '11 at 8:09
In practical situations we estimate $\sigma^2$, then $\hat d$ is no longer gaussian. Instead, it is a $t$. But a $t$ with higher than 10 is not very different from a gaussian distribution. –  suncoolsu Jul 13 '11 at 14:31
@suncoolsu: I absolutely don't understand what you mean :) Please clarify and give a reference. Thank you! –  MarkDollar Jul 13 '11 at 17:27

The $u$s are unobserved and the $\hat{d}$s are just estimates of them.