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Let X have a density function $f_X(x) = x^3/4$ for 0 < x < 2, otherwise $f_X(x) = 0$

Let $Z = \sqrt{X}$. Compute the density function of $f_Z(z)$ for $Z$.


My attempt:

Let $h(x) = \sqrt{x}$, then $h^{-1}(z) = z^2$, and the derivative

$h'(x) = \frac{1}{2x^{1/2}}$

$$f_Z(z) = \frac{f_X(z^2)}{1/2z} = \frac{z^6/4}{1/2z} = z^7/2$$ for $0 < z < \sqrt{2}$, $0$ otherwise.


The solution in my textbook:

Since $Z^{-1}(z) = z$ and $Z'(z) = 1$, $f_Z(z) = f_X(z)(1) = z^3/4$.

Why is my answer wrong?

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  • $\begingroup$ What is the title of your textbook? $\endgroup$ – Did Jul 22 '18 at 8:43
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I agree with your answer. let $z \in (0,\sqrt2)$, \begin{align} Pr(Z \le z) &= Pr(\sqrt{X} \le z)\\ &= Pr(X \le z^2)\\ &=\int_0^{z^2}f_X(t)\, dt \end{align}

$$f_Z(z) = \frac{d}{dz} \int_0^{z^2}f_X(t) \, dt=f_X(z^2) (2z)=\frac{z^2}{4}(2z)=\frac{z^7}2$$

I don't see anywhere in the proposed solution that they use the property that $Z=\sqrt{X}$ at all.

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Your answer is correct. An alternative way to verify is to note that $$\Pr\{Z \le z\} = \Pr\{\sqrt{X} \le z\} = \Pr\{X \le z^2\} = F_X(z^2).$$ Therefore, $$f_Z(z) = \frac{d}{dz} F_X(z^2) = 2zf_X(z^2).$$

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