Let $X$ be a random variable which distribution function is defined as $F_x(x):=\frac{(x+1)^2}{4} \forall x\in(-1,1)$

Let $Y:=g(X)$ where $g(x):=5x+7$

Would this reasoning be correct?

$F_y(y)=Pr(Y\leq y)=Pr(g(X)\leq y)=Pr(5X+7\leq y)=Pr(X\leq\frac{y-7}{5})=F_x(\frac{y-7}{5})=\frac{(\frac{y-7}{5}+1)^2}{4}=\frac{y^2-4y+4}{100} \forall y\in(-1,1)$

I have doubts whether the interval is actually $(-1,1)$ as well, because in discrete changes of variable, those can change. That doesn't apply to continuous variables as well?

I am also concerned by the fact that $F_y$ is not and increasing function, which all distribution functions should, how is this?

Also, the resulting distribution $F_y$ is not continuous, as $$F_y(-1)^{-}=0, F_y(-1)^{+}=\frac{9}{100}$$ and $$F_y(1)^{-}=\frac{1}{100}; F_y(1)^{+}=1$$

Any thoughts?


The formula you obtained for $F_Y(y)$ is valid for $2\le y\le 12$. On this interval, the function is nicely behaved. For completeness, observe that $F_Y(y)=0$ for $y\lt 0$, and $F_Y(y)=1$ for $Y\gt 12$.

  • $\begingroup$ Oh true, thank you André, I didn't realize you should also adjust the interval itself, that does indeed solve everything :) $\endgroup$ – F.Webber May 12 '14 at 20:23
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    $\begingroup$ You are welcome. Sure, $Y$ does not "live" on $[-1,1]$. $\endgroup$ – André Nicolas May 12 '14 at 20:31
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    $\begingroup$ Actually $F_Y(y)=1$ for $y\geqslant12$, not for $y\gt1$. $\endgroup$ – Did May 12 '14 at 21:18
  • $\begingroup$ I have one more doubt closely related to this problem, what if we defined g as $g(x):=x^2$, and wanted, again, to find the new distribution function $F_y$. Now I'm having problems with that one. The resulting expression is $\frac{(\sqrt{y}+1)^2}{4}$, but, which is the interval that is valid in? How to find it? $\endgroup$ – F.Webber May 12 '14 at 21:47
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    $\begingroup$ @LMartin: If $Y=X^2$ then $Y$ lives on $[0,1]$. We have for $y$ in this interval $\Pr(Y\le y)=\Pr(-\sqrt{y}\le X\le \sqrt{y}=F_X(\sqrt{y})-F_X(-\sqrt{y})$. $\endgroup$ – André Nicolas May 12 '14 at 22:10

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