I have a standard exponential CDF:
$F_X(x) = $ \begin{cases} 1-e^{\lambda x}, & \text{x $\ge$ 0} \\ 0, & \text{otherwise} \end{cases}
$ Y = ((X-2)^2-1)^2\\X = 2\pm(1\pm \sqrt{Y})^{1/2} $
If someone could please edit my equation to show + or - better, thank you.
I want to write the CDF of
$F_Y(y) = ?$
This is not straight forward because multiple values of Y map into the same value of X. (I think I'm saying that right) Therefore, I have to add the probabilities for instances where one value of Y, lets say 0.5, will produce 4 different values depending on if you are taking plus/minus. I'm not sure if I'm thinking about this right however, this is what I have managed to pierce together:
$F_Y(y) = $
\begin{cases} 4-e^{\lambda x_1}-e^{\lambda x_2}-e^{\lambda x_3}-e^{\lambda x_4}, & \text{0 $\le$ Y $\le$1} \\ 2-e^{\lambda x_1}-e^{\lambda x_3}, & \text{Y > 1} \\ 0, & \text{otherwise} \end{cases}
Where $x_i$ is equal too:
$x_1 = 2+(1+\sqrt{y})^{1/2}\\x_2 = 2+(1-\sqrt{y})^{1/2}\\x_3 = 2-(1+\sqrt{y})^{1/2}\\x_4 = 2-(1-\sqrt{y})^{1/2}$
In other words, I'm trying to add the intervals up where the Y variable exists for values of X. Since between [0,1], y can map into four possible x values, I need to add those four mappings up in the CDF. I would appericate some help. TY
In response to some comments, I am posting further work I have done.
Plotting the function g(x):
The red line represents y and all of the graph below that line is where the inequality has to be written for: ( I am trying to show between points #1,#2 and #3 & #4.
$ 2-(1+\sqrt{y})^{1/2}\le x \le 2-(1-\sqrt{y})^{1/2}\\2+(1-\sqrt{y})^{1/2}\le x \le 2+(1+\sqrt{y})^{1/2}$
Are these the two inequalities that show the areas where x is less than some y?
$ if y > 1 \\P(2-(1+\sqrt{y})^{1/2}\le x \le 2+(1+\sqrt{y})^{1/2}$