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Let $Y$ be a random with the following probability density :

$$ f_Y(y) = \begin{cases} 2y, & 0 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases} $$

Let $U = 3Y -1$. I want to find the density function $f_U(u)$. I know I can find it using the Method of Distribution Functions namely by finding $F_U(u) = P(U \leq u) = P(Y \leq \frac{u+1}{3})$ and then differentiating $F_U(u)$. However, I wonder why it is not possible to directly calculate $f_U(u)$ as follows :

$$f_U(u) = P(U = u) = P(Y = \frac{u+1}{3})$$

It does give the same result. Why?

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    because it is not true that $f_U(u)= P(U=u)$. Note that since $Y$ is continuous $U$ is continuous and hence $P(U=u)=0$ for all $u \in \Bbb R$. – 2017-01-12
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    Makes sense! Thanks :) – 2017-01-12
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    @Dory: I tried to apply your intuitive approach and to give a little more precise solution. Please, take a look. – 2017-01-12

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The OP's idea can be saved the following way. Take a small $\Delta u>0$ and then

$$f_U(u)\Delta u\approx P(u\leq U

Dividing both sides by $\Delta u$ we get that

$$f_U(u)\approx\frac{F_Y\left(\frac{u+1}3+\frac{\Delta u}{3}\right)-F_Y\left(\frac{u+1}{3}\right)}{\Delta u}.$$

Letting $\Delta u\to 0$ it turns out that

$$f_U(u)=\frac1{3} f_Y\left(\frac{u+1}3\right)=\frac2{9}\times\begin{cases} u+1&\text{ if }& -1\leq u \leq 2\\ 0&\text{ otherwise. } \end{cases}$$

Note

I am not saying that this is the way such problems have to be solved...

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    If you're not saying, then I will: This is exactly *the* way such problems should be solved. +1 – 2017-01-12
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    @zhoraster: Thanks for you remark. Yet, in my opinion, the best is the "distribution function method" for it is not dependent on the existence of the densities. The worst method is, of course, to learn the magic formula which exists in the case of existing densities. – 2017-01-13
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    Quite clever way of doing it! :) @zhoraster -1. There are multiple ways to solve such problems using for instance inverse functions, jacobians or even moment functions. – 2017-01-13
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    @Dory: I don't quite understand your comment above. (But, to be honest, as a fierce point collector, I would like to know if you accept my solution or you don'T.) – 2017-01-13
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    @zoli, the magic formula is worst only in the case if it is learned and applied mechanically. Otherwise, I think it is much better than the cdf method, which requires solving *inequalities*. – 2017-01-13
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    @zhoraster: I agree. Sorry, if my remark could be misunderstood. – 2017-01-13