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I have the following PDF that gives the probability of a certain annual wage being drawn:

$f(w)=0$ if $w<20000$

$\frac{w-20000}{50000^2}$ if $w \in [20000,70000]$

$\frac{120000-w}{50000^2}$ if $w \in (70000,120000]$

$0$ if $w>120000$

I want to scale $w$ to $w'=\frac{w}{365}$, so that I can analyze daily wage instead. How do I scale the pdf accordingly?

There are my initial thoughts:

$f(w')=0$ if $w'<\frac{20000}{365}$

$\frac{w'-\frac{20000}{365}}{(\frac{50000}{365})^2}$ if $w'\in [\frac{20000}{365},\frac{70000}{365}]$

$\frac{\frac{120000}{365}-w'}{(\frac{50000}{365})^2}$ if $w' \in (\frac{70000}{365},\frac{120000}{365}]$

$=0$ if $w'>\frac{120000}{365}$

Is this correct?

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    $\begingroup$ The pdf of the annual wages is a triangle shape with base of length 100,000 and maximum height $(50,000)^{-1}$ for a total area of $1$. The pdf of the daily wage is also a triangle, but the base length is smaller by a factor of 365, and so the height must be larger by the same factor to keep the area the same. Check if this is what your formula gives you, and if it does, you have answered your own question! $\endgroup$
    – Dilip Sarwate
    Dec 20, 2013 at 4:12

1 Answer 1

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It seems correct but in general if you have a function that is 1-to-1, say $h$, and $Y=h(X)$ then the pdf of Y is related to the pdf of X by the change of variables formula:

$$ f_Y(y)= f_X\left(h^{-1}(y)\right)\cdot\left|\dfrac{\partial}{\partial y}h^{-1}(y)\right|$$ and obviously if $X$ has support on $[a,b]$ then $Y$ has support on $[h(a),h(b)]$.

In your case $h(x)=\frac{x}{365}$ so $h^{-1}(y)=365\cdot y$ and $\left|\dfrac{\partial}{\partial y}h^{-1}(y)\right|=365$.

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