Simple proof on uniform distribution

Question

We have two random variables, $X$ and $Y$. $X$ follows uniform distribution. If $Y = aX + b$, how is any particular percentile of the Y distribution related to the corresponding percentile of the X distribution?

I know the answer is straight forward: If the value of $m$ has a particular percentile in $X$, the value of $am+b$ has the same percentile in $Y$. However, I just do not know how to prove this.

Can't you use the fact that $ F_Y (y) = {Y< y } $ then $ F_X (y) = { aX+b < y } = { X <(y-b)/a } $ ? — Malcolm, Mar 11 '16 at 08:19
I think that the question is to prove $F_Y(am+b) = F_X(m)$, for all $m \in \mattabb{R}$? — Joseph, Mar 11 '16 at 08:25
What I gave you should give you enough information, simply replace y with am+b — Malcolm, Mar 11 '16 at 08:30
I am not the brightest in stats. So you mean: $F_Y(y) = P(Y<y) = P(aX+b<y) = P(X<\frac{y-b}{a}) = F_X(\frac{y-b}{a})$? — Joseph, Mar 11 '16 at 08:40
Oh, so the solution is: $F_X(x) = P(X < x) = P(\frac{Y-b}{a} < x) = P(Y<ax+b) = F_Y(ax+b)$ — Joseph, Mar 11 '16 at 08:45
Since $Y=aX+b$ then you can say that $ y = am+b $ for all values of the random variable, thus from what we said before, $F_Y(y)=P{ Y < y } = P{aX+b<y } = P{X< \frac{y-b}{a} } = F_X(\frac{y-b}{a})$ Since y=am+b, then $F_X(\frac{y-b}{a})=F_X(m) $ Thus you've shown that $F_Y(y)=F_Y(am+b)=F_X(m)$ — Malcolm, Mar 11 '16 at 08:50

score 1 · Answer 1 · answered Mar 11 '16 at 08:45

1

Let $m$ the $\alpha$-th percentile of $X$. By definition, we have $P(X\leq m)=\alpha$.

So if $a>0$, $\alpha=P(X\leq m)=P(aX+b\leq am+b)=P(Y\leq am+b)$

We have $P(Y\leq am+b)=\alpha$, which means that $am+b$ is the $\alpha$-th percentile of $Y$, by definition.

answered Mar 11 '16 at 08:45

Augustin

8,446

So you mean: $F_X(x) = P(X < x) = P(aX+b < ax+b) = P(Y<ax+b) = F_Y(ax+b)$. Am I right? – Joseph Mar 11 '16 at 08:50
But your method only work if $a \geq 0$? If $a < 0$, then the sign $\leq$ will change? – Joseph Mar 11 '16 at 08:52
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Yes, if $a<0$ you have to change the inequality. But then the answer is no longer $am+b$. – Augustin Mar 11 '16 at 08:58
If$a < 0$, then $F_X(m) = P(X < m) = P(aX+b > am+b) = P(Y > am+b)$. And then? – Joseph Mar 11 '16 at 09:02
@Joseph $P(Y>am+b)=1-P(Y\leq am+b)$ So if $m$ is the $\alpha$ percentile of $X$, what is $am+b$ to $Y$? – Graham Kemp Mar 11 '16 at 09:05
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@GrahamKemp Search me. – Joseph Mar 11 '16 at 09:06
If $a<0$, as $F_X(m) = P(Y > am+b) = 1-P(Y \leq am+b)$, then $P(Y \leq am+b) = 1 - F_X(m)$ – Joseph Mar 12 '16 at 13:48

Graham Kemp · Answer 2 · 2016-03-11T08:57:43.700

If $X\sim\mathcal U(\alpha;\beta)$ then $F_X(x)=\dfrac{x-\alpha}{\beta-\alpha}~\raise{0.25ex}{\chi}_{x\in(\alpha;\beta)}+\raise{0.25ex}{\chi}_{x\in[\beta;\infty)}$

The percentile function is the inverse of the CDF.

$$\begin{align}Q_X(p) = & ~ F_X^{-1}(p/100) \\[1ex] = & ~ \alpha+(\beta-\alpha) p/100\end{align}$$

Since $Y=aX+b$ then $Y\sim\mathcal U(a\alpha+b, a\beta+b)$ (If $a>0$)

$$\begin{align}Q_Y(p) = & ~ F_Y^{-1}(p/100) \\[1ex] = & ~ a\alpha+b+(a\beta+b-a\alpha-b)p/100 \\[1ex] = & ~ b+a(\alpha+(\beta-\alpha)p/100) \\[1ex] = & ~ a~Q_X(p)+b\end{align}$$

Which is as was anticipated.

Can you do $a<0$?

Simple proof on uniform distribution

2 Answers2