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I'm given an absolutely continuous stochastic variable $U$, with the probability density function $f_{U}$ given by; $$ f_{U}\left(u\right) = \left\{ \begin{array}{ll} 2u & \text{ if } u \in \left] 0,1 \right[ \\ 0 & \text{ otherwise.} \end{array} \right. $$

My task is now to derive the probability density function for $X = e^{U}$. I've tried different approaches, however I'm unable to get the expected result of; $$ f_{X}\left(x\right) = \left\{ \begin{array}{ll} \frac{2 \cdot \ln\left(x\right)}{x} & \text{ if } x \in \left] 1,e \right[ \\ 0 & \text{ otherwise.} \end{array} \right. $$

Edit:

Found another exercise alike this one (which seems simpler)

I'm given an stochastic variable $Y \sim R\left(0,1\right)$, with the probability density function $f_{Y}$ given by; $$ f_{Y}\left(y\right) = \left\{ \begin{array}{ll} 1 & \text{ if } y \in \left]0,1\right[ \\ 0 & \text{ otherwise.} \end{array} \right. $$

My task is now to derive the probability density function for $Z = Y^{3}$. Again without any luck I've tried different approaches, and I'm unable to reach the expected result of; $$ f_{Z}\left(z\right) = \left\{ \begin{array}{ll} \frac{1}{3} \cdot z^{-\frac{2}{3}} & \text{ if } z \in \left]0,1\right[ \\ 0 & \text{ otherwise.} \end{array} \right. $$

Note

So far I've been able to determine the correct intervals, using the method @DilipSarwate.

After I figured, that I was integrating, rather than differentiating (thanks to @JonathanY.), I'm able to get the first exercise, to get the correct answer, the second is however troublesome:

Differentiating $z^{\frac{1}{3}}$ yields $\frac{1}{3} \cdot z^{\frac{2}{3}}$, and not; $\frac{1}{3} \cdot z^{-\frac{2}{3}}$

Question

What should I do in order to reach the correct result (please see discussion on the answer)?

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  • $\begingroup$ Differentiating $z^{\frac{1}{3}}$ yields $\frac{1}{3} \cdot z^{\frac{2}{3}}$, and not $\frac{1}{3} \cdot z^{-\frac{2}{3}}$. True or False: For $a \neq 0$, the derivative of $x^a$ is $ax^{a-1}$ ?? $\endgroup$ – Dilip Sarwate Aug 12 '13 at 2:10
  • $\begingroup$ The last statement is true, and I do believe that the differential of $z^{\frac{1}{3}$ is correct, but if this is the case, then my expected result, must be wrong, can you confirm this? $\endgroup$ – Skeen Aug 12 '13 at 13:24
  • $\begingroup$ Another Question: Does $\frac{1}{3} - 1$ equal $\frac{2}{3}$ or does it equal $-\frac{2}{3}$? If the latter, what does setting $a = \frac{1}{3}$ in the statement $$\frac{d}{dz}z^{a} = az^{a-1}$$ yield as the derivative of $z^{\frac{1}{3}}?\ \ $ $\frac{1}{3}z^{-\frac{2}{3}}$ or $\frac{1}{3}z^{\frac{2}{3}}$ as you have repeatedly insisted? $\endgroup$ – Dilip Sarwate Aug 12 '13 at 22:31
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First, figure out what values $X = e^U$ can take on in view of the given information about $U$.

Next, write $$F_X(x) = P\{X \leq x\} = P\{e^U \leq x\} = P\{U \leq g(x)\} = F_U(g(x))$$ where you should figure out what $g(x)$ is.

Then, write

$$f_X(x) = \frac{\mathrm d}{\mathrm dx}F_X(x) = \frac{\mathrm d}{\mathrm dx}F_U(g(x)) = f_U(g(x))\frac{\mathrm d}{\mathrm dx}g(x)$$ where once again you need to figure out the derivative of $g(x)$. If the last equality confuses you recall the chain rule for differentiation from basic calculus.


Applying the idea above, if $X$ is uniformly distributed on $(0,1)$, then $Y = X^3$ also takes on values in $(0,1)$ and so for $0 < y < 1$, $Y$ has distribution function $$F_Y(y) = P\{Y \leq y\} = P\{X^3 \leq y\} = P\{X \leq y^\frac{1}{3}\} = F_X(y^\frac{1}{3})$$ and so, $$f_Y(y) = \frac{\mathrm d}{\mathrm dy}F_Y(y) = \frac{\mathrm d}{\mathrm dy} F_X(y^\frac{1}{3}) = f_X(y^\frac{1}{3})\cdot \frac{1}{3}y^{-\frac{2}{3}} = \frac{1}{3}y^{-\frac{2}{3}}, 0 < y < 1.$$ Here we have used the chain rule for differentiation and the facts that $f_X(\cdot)$ has constant value $1$ for arguments in $(0,1)$, and that for $a \neq 0$, $\frac{\mathrm d}{\mathrm dy}y^a = ay^{a-1} = \frac{1}{3}y^{-\frac{2}{3}}$ when $a = \frac{1}{3}$.

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  • $\begingroup$ <quote>First, figure out what values $X = e^U$ can take on in view of the given information about $U$.</quote> I've already tried this, by setting $u$ to it's limits (0, and 1), which yields the limits $1$ and $e^{2}$ for $x$ ($e^{2} \neq e$)? $\endgroup$ – Skeen Aug 9 '13 at 12:43
  • $\begingroup$ In ordinary mathematical discourse, the symbol $e$, unless explicitly defined to mean something else, usually means the base of the natural logarithms; and so $e^2 \neq e$. Also, I don't understand how you get that $e^U = e^2$ when $U = 1$. Could you explain further? $\endgroup$ – Dilip Sarwate Aug 9 '13 at 13:16
  • $\begingroup$ The $e$ is indeed the base of natural logarithms. And as for why I got $e^{2}$, that's because I've translated $e^{U}$ to $e^{2*u}$ and then used my values as $u$, should I be using my values for $e^{U}$? $\endgroup$ – Skeen Aug 9 '13 at 16:16
  • $\begingroup$ @Skeen $X = e^U$ and so if $U = 1$, $X = e$, no? Why should the result be any different if you are given additional information to distract you (such as density function of $U$)? $\endgroup$ – Dilip Sarwate Aug 10 '13 at 4:20
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    $\begingroup$ You've confused derivative with anti-derivative. Differentiate $\ln(x)$ to get $\frac{1}{x}$. $\endgroup$ – Jonathan Y. Aug 11 '13 at 20:43

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