So this was my solution:

Say, $Z = X^{2}$, then $X=\pm \sqrt{Z}$


$$P(Z=z)=P(X = \sqrt{z}) + P(X = -\sqrt{z}) = \frac{z}{18} + \frac{z}{18} = \frac{z}{9}$$




$$\int_{0}^{9}P(z) dz = \int_{0}^{9}\frac{z}{9}dz = \frac{9}{2} \neq 1$$

Where did I go wrong?

Edit: I tried to do this problem intuitively. I looked through my textbook and found:

$$f_{Y}(y) = f_{X}(g^{-1}(y))\left | \frac{dx}{dy} \right |$$

So updated question: why do we need to multiply by the jacobian? Is it because if you were to do this problem using the cdf and took its derivative to get the pdf, you would end up with that term which happens to be the jacobian? Is there a way to intuitvely understand why that term is there?



with a pdf, the probability of the random variable to be equal to a given value is $0$.

The correct answer is $g(z)=\dfrac{\sqrt{z}}{18}$ for $0\leq z<9$, proof :

With $Z=X^2$, and $g$ the pdf of $Z$:

$$P(0\leq Z < a) = P(-\sqrt{a}<X<\sqrt{a}) = \int_{-\sqrt{a}}^{\sqrt{a}} \frac{x^2}{18}dx$$

given that :

$$\int_{-\sqrt{a}}^{\sqrt{a}} \frac{x^2}{18}dx = \frac{a^{\frac{3}{2}}}{27}$$

and $$P(0\leq Z < a) = \int_0^a g(z)dz$$

We have $$\int_0^a g(z)dz = \frac{a^{\frac{3}{2}}}{27}$$

and from that we get $g(z) = \dfrac{\sqrt{z}}{18}$ with the derivative


$P(X^2\le x)=P(-\sqrt{x}\le X\le \sqrt{x})=\int_{-\sqrt{x}}^\sqrt{x}\frac{x^2}{18} dx=\frac{x^{1.5}}{27}\ 0\le x\le 9$

$f_{X^2}(x)=\frac{x^{0.5}}{18}\ 0\le x\le 9$


Just use the formula for the function of rv and: $$ X=\pm \sqrt{Z}\\ \frac{dX}{dZ} =\pm \frac{1}{2 \sqrt{Z}} $$ Hence, $$ f_{Z}(z)=\frac{z}{18} \cdot \frac{1}{2 \sqrt{z}} + \frac{z}{18} \cdot \frac{1}{2 \sqrt{z}} = \frac{\sqrt{z}}{18}, \ 0 \leq z \leq 9 $$

  • $\begingroup$ Yup. Is there an intuitive explanation though, as to why we need to multiply by the Jacobian? I am thinking it has to do with sets: if I take the difference between 1.01^2 and 1.02^2 the result is more than .01 (which is the difference between the inputs). So when X is squared, the domain of X^2 is somehow not the continuous number line? And the Jacobian is required to correct for that? Is this along the right tracks? $\endgroup$ – kyphos Mar 6 '13 at 0:05
  • $\begingroup$ because $f(x)=F'(x)$ $\endgroup$ – Alex Mar 6 '13 at 0:52

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