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Given $X\sim U(0,1)$, i have to determine the density of $Y=-\frac{1}{\lambda}lnx$.

I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:

$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(-lnX\leq \lambda y)=\mathbb{P}(X\geq e^{-\lambda y})=1-\mathbb{P}(X\leq e^{-\lambda y})=1-F_X(e^{-\lambda y})$

Now it's clear that $f_Y(y)=\lambda e^{-\lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-\lambda y})$ and $f_Y(y)=\lambda e^{-\lambda y}$. Anyone can help me?

Thanks in advance!

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  • $\begingroup$ Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation. $\endgroup$ – Marco Pittella Dec 11 '18 at 10:41
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Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.

Go one step further in the calculation and write: $$F_Y (y)=1-e^{-\lambda y} $$

This is allowed because $F_X (x)=x $ for $x\in (0,1) $.

Now take the derivative of $F_Y (y) $ and you are ready.

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  • $\begingroup$ Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x \in [0,1]$, so $F_Y(y)=1-F_X(e^{-\lambda y})=1-e^{-\lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Y\notin \mathbb{R}$. Moreover, if $X\in [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-\lambda y}\in [0,1]$. But i don't understand how this help me to formalize the passage above. $\endgroup$ – Marco Pittella Dec 10 '18 at 7:07
  • $\begingroup$ I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-\lambda y})$ for $y>0$? 2) Do you agree that $e^{-\lambda y}\in(0,1)$ for $y>0$ and $\lambda>0$? 3) Do you agree that $F_X(x)=x$ for $x\in(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-\lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $y\leq0$ so the complete CDF of $Y$ has been found. $\endgroup$ – drhab Dec 10 '18 at 9:19
  • $\begingroup$ 1) Why $y>0$? 2) Why $e^{-\lambda y}\in (0,1)$ $\endgroup$ – Marco Pittella Dec 10 '18 at 15:51
  • $\begingroup$ Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-\frac1{\lambda}\ln X$ only takes values in $(0,\infty)$. That makes it clear immediately that $F_Y(y)=P(Y\leq y)=0$ for $y\leq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-\lambda y}\in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment. $\endgroup$ – drhab Dec 10 '18 at 18:16
  • $\begingroup$ Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-\lambda y}\leq 1$, i can write $P(X\geq e^{-\lambda y})=\int_{e^{-\lambda y}}^{\infty}f_X(x)dx=\int_{e^{-\lambda y}}^{1}1dx+\int_{1}^{\infty}0dx=\int_{e^{-\lambda y}}^{1}1dx=[x]_{e^{-\lambda y}}^{1}=1-e^{-\lambda y}$ or $1-P(X\leq e^{-\lambda y})=1-\int_{-\infty}^{e^{-\lambda y}}f_X(x)dx=1-[\int_{-\infty}^{0}0dx+\int_{0}^{e^{-\lambda y}}1dx]=1-[x]_{0}^{e^{-\lambda y}}=1-e^{-\lambda y}$. $\endgroup$ – Marco Pittella Dec 11 '18 at 10:39

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