# Why is there a linear relationship between the quantiles of a generic normal distribution and those of the standard normal distribution?

I don't understand why a quantile-quantile plot is linear. That is, if you plot quantiles of the normal distribution with mean $\mu$ and standard deviation $\sigma$ (on the vertical axis) against quantiles of the standard normal distribution (on the horizontal axis), you get a line. Can someone gently explain the linear relationship? What is the slope of that line?

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Let $y_t$ be the $t$-th quantile of $Y$, where $Y$ is normal with mean $\mu$ and variance $\sigma^2$. So $\Pr(X\le y_t)=t$.

Let $X$ be standard normal, and let $x_t$ be the $t$-th quantile of $X$. We have $$t=\Pr(Y\le y_t)=\Pr\left(\frac{Y-\mu}{\sigma}\le \frac{y_t-\mu}{\sigma}\right)=\Pr\left(X\le \frac{y_t-\mu}{\sigma}\right).$$ It follows that $\dfrac{y_t-\mu}{\sigma}=x_t$.

We can rewrite this as $y_t=\sigma x_t +\mu$.

Thus whatever $t$ is, the point $(x_t,y_t)$ lies on the line with equation $y=\sigma x+\mu$. We recognize this as the equation of a line with slope $\sigma$.

Remark: We gave a formal argument. But informally, the result is intuitively almost clear. A general normal is just a standard normal scaled then shifted, just like temperature in degrees Fahrenheit is temperature in degrees Celsius scaled and shifted. More precisely, $Y=\sigma X+\mu$.

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Thank you, I'm trying to parse your formal argument. I agree completely that this is intuitive, which makes it all the more frustrating that I can't prove it to myself. What you wrote is still a bit unclear to me, but I will stare at it and see if it begins to make sense. –  user7661 Jan 27 '13 at 3:24
Yes, lots of symbols. If you like, you can replace $y_t$ and $x_t$ by $y$ and $x$. I used $X$ for the standard normal instead of the usual $Z$. That was to get $x$-axis, $y$-axis as in your post. It all comes down to how we compute probabilities for a general normal by using the standard normal. –  André Nicolas Jan 27 '13 at 3:38