1
$\begingroup$

I am reading some probability courses on the https://www.probabilitycourse.com/chapter4/4_1_3_functions_continuous_var.php . It says the following :

Let X be a Uniform(0,1) random variable, and let $Y=e^X$

enter image description here

I would like to ask why the following equation is correct:

$$P(e^X \leq y)=P(X \leq \ln y)$$

Intuitively, we can apply the ln operator in the first CDF to get the second CDF. However, I am confused when I expand these 2 cumulative distribution functions into the integral format.

$$P(Y \leq y) = P(e^X \leq y)=\int_{{-\infty}}^{y} f_y(y) dy=\int_{{-\infty}}^{y} f_y(e^x) dy=\int_{{-\infty}}^{\ln y} f_y(e^x) e^x dx$$

$$P(X \leq \ln y)=\int_{{-\infty}}^{\ln y} f_x(x) dx$$

$f_x$ and $f_y$ are the corresponding pdfs.

I can not see directly that the equation $\int_{{-\infty}}^{\ln y} f_y(e^x) e^x dx = \int_{{-\infty}}^{\ln y} f_x(x) dx$ can really hold .

So I would like to ask why the transformation in the scope of the cumulative distribution function works ? Did I miss some axiom ? I guess it should be a basic question but I can not get it from google.

Thank you


I agree with @Grada Gukovi's comment, and I find the Method of Transformation will produce the equation $ f_y(e^x) e^x = f_y(y) e^x = f_x(x) $ .

What I think in this problem is that , the statement of $P(e^X \leq y)=P(X \leq \ln y)$ should not be a simple transformation without evidence. Because $P(e^X \leq y)$ is the integral for random variable y and $P(X \leq \ln y)$ is the integral for random variable x. This equation is not easy to be derived and gives me lots of worrys. From my thinking, we can only get this equation after using the Method of Transformation.

I guess we can not say $P(e^X \leq y) = P(X \leq \ln y) $ if $(e^X \leq y) = (X \leq \ln y) $ simply.

$\endgroup$
2
$\begingroup$

The example you are asking about is a demonstration of the Method of Transformations that is described further down the page. As the example says $X$ is distributed as $Unif(0,1)$ and $Y = e^x$. Thus the probability of Y being smaller or equal than some value $z$ $P(y \leq z) = P(e^x \leq z)$. (i)

Since the exponential function is strictly increasing and maps $[0,1] \rightarrow [e^0,e^1]$ it has an inverse function that maps $[e^0,e^1] \rightarrow [0,1]$ (the natural logarithm). So we have $e^x \leq z \Leftrightarrow x \leq log(z)$. And therefore $P(e^x \leq z) = P(x \leq log(z))$. (ii)

Combining (i) and (ii) gives $P(y \leq z) = P(e^x \leq z) = P(x \leq log(z)) $.

As far as your integration is concerned it is:

$\int_{{-\infty}}^{y} f_y(e^x) dy=\int_{{-\infty}}^{\ln y} f_y(e^x) e^x dx = \int_{{-\infty}}^{\ln y} f_x(e^x)dx$,

by applying the Method of Transformation to $f_y$ to transfrom it to $f_x \times \frac{1}{e^x}$.

$\endgroup$
  • $\begingroup$ Thank you. I agree with your comment, and I find the Method of Transformation will produce the equation $ f_y(e^x) e^x = f_y(y) e^x = f_x(x) $ . What I think in this problem is that , the statement of $P(e^X \leq y)=P(X \leq \ln y)$ should not be a simple transformation without evidence. Because $P(e^X \leq y)$ is the integral for random variable y and $P(X \leq \ln y)$ is the integral for random variable x. This equation is not easy to be derived and gives me lots of worrys. From my thinking, we can only get this equation after using the Method of Transformation. $\endgroup$ – zhfkt Aug 7 at 12:36
  • $\begingroup$ In short, I guess we can not say $P(e^X \leq y) = P(X \leq \ln y) $ if $(e^X \leq y) = (X \leq \ln y) $ simply. $\endgroup$ – zhfkt Aug 7 at 12:47
2
$\begingroup$

The reason $e^X\le y,\,X\le\ln y$ have the same probabilities of being true is because they're equivalent statements, which in turn is because $y=e^x$ is strictly increasing, with inverse $x=\ln y$.

$\endgroup$
  • $\begingroup$ Yes. From my thinking, this equation is not easy to be derived and gives me lots of worrys. Because $P(e^X \leq y)$ is the integral for random variable y and $P(X \leq \ln y)$ is the integral for random variable x. I guess we can not say $P(e^X \leq y) = P(X \leq \ln y) $ if $(e^X \leq y) = (X \leq \ln y) $ simply. $\endgroup$ – zhfkt Aug 7 at 12:40
  • 2
    $\begingroup$ @zhfkt There's no problem here: $((e^X\le y)\iff(X\le\ln y))\implies P(e^X\le y)=P(X\le\ln y)$. $\endgroup$ – J.G. Aug 7 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.