PDF of function of uniform random variable

Question

Why PDF of $g(X)=X^3$ is not uniformly distributed, when X is uniform random variable between $(0,1)$? As for every value of X there is unique value of $g(X)$, hence the probability density of $g(X)$ should be equal to probability density of X?

Answer 1

Since $X\sim \operatorname{Uniform}(0,1)$ we know that $F_X(x) = \mathbb{P}(X \leq x) = x$. Let $Y = X^3$. $$F_Y(y) = \mathbb{P}(X^3 \leq y) = \mathbb{P} (X \leq y^{1 \over 3}) = y^{1 \over 3}.$$ Taking the derivative of the cumulative distribution function we obtain the pdf: $$F'_Y(y) = f_Y(y)= {1\over 3} y^{- {2\over 3}}.$$ We can see from the pdf that $Y$ is not uniformly distributed.

Answer 2

If $f$ denotes the probability of some random variable then this does not mean that $f(x)=P(X=x)$ for each $x$. It seems that you are thinking this way.

Yes. In the context of your question: if $y=x^3$ then we have $P(X^3=y)=P(X=x)=0$.

But that is not a determining relation if it comes to the PDF of $X^3$.

Answer 3

Because you have that

$$P( X < \frac{1}{2} ) = \frac{1}{2}$$

and

$$P( X^3 < \frac{1}{2} ) = P( X < \frac{1}{\sqrt[3]{2} } ) = \frac{1}{\sqrt[3]{2} }$$

And both are clearly different. So the repartition functions are differents. But the density is the anti-derivative of the repartition function, so these are different too.

Answer 4

It is true that $f(x)=x^3$ pairs off each member of $[0,1]$ with exactly one other member of $[0,1]$, i.e. this function is a bijection from $[0,1]$ to itself. But continuous probability distributions don't really care about this. (Discrete distributions do care about this.) Instead, if $X$ is uniform on $[0,1]$, then $X^3$ is more likely to be small than it is to be large. For instance, $X^3 \leq 0.008$ if and only if $X \leq 0.2$, so the probability of that event is $0.2$, not $0.008$, as it would be if $X^3$ were uniform.

This might be easier to see in a more extreme example: pick some huge positive number $M$ and take $X$ uniformly distributed on $[-M,0]$. Then consider $Y=e^X$. You again have a bijection, in this case onto $[e^{-M},1]$, but $Y$ will not be uniform. Instead it will almost always be quite small.