# Proof or relation between a Uniform and Exponential

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?

• Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation. – Marco Pittella Dec 11 '18 at 10:41

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.

• 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. – Marco Pittella Dec 10 '18 at 7:07
• 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. – drhab Dec 10 '18 at 9:19
• 1) Why $y>0$? 2) Why $e^{-\lambda y}\in (0,1)$ – Marco Pittella Dec 10 '18 at 15:51
• 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. – drhab Dec 10 '18 at 18:16
• 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 $1-P(X\leq e^{-\lambda y})=1-\int_{-\infty}^{e^{-\lambda y}}f_X(x)dx=1-[\int_{-\infty}^{0}0dx+\int_{0}^{e^{-\lambda y}}1dx]=1-[x]_{0}^{e^{-\lambda y}}=1-e^{-\lambda y}$. – Marco Pittella Dec 11 '18 at 10:39