Probability density function from known density function

Question

Suppose I want to compute new density function $Z$, $Z= min(X_1,2X_1-X_2)$, where $X_1$ and $X_2$ are independent exponential distributions with $\lambda$. I have looked into Finding the density for $\min\{X, Y\}$ and in their case, $X, Y $ are independent, so they can split the cdf into a product. But here, I don't think $X_1,2X_1-X_2$ are independent because of the $2X_1-X_2$ How would I tackle this? I also know that the sum of two exponential distributions is a gamma distribution but I don't know if they are independent so I cannot proceed. Any help is appreciated!

Answer 1

Observing that $3>2$ and $X_1,X_2 \geq 0$ it results to me that

$$3X_1\geq 2X_1-X_2$$

$\forall x_1,x_2 \in \mathbb{R}^+$

Thus

$$Z=2X_1-X_2$$

With usual transformation procedures the distribution of Z can be derived without particular problems

---

EDIT:

Let's use the definition:

$$ \bbox[yellow,5px,border:2px solid red] { \mathbb{P}[Z>z]=\mathbb{P}[min(X,2X-Y)>z]=\mathbb{P}[X>z;Y<2X-z] \ } $$

Do a drawing and derive the region to integrate $f_{XY}(x,y)$ to find $1-F_Z(z)$

---

sketch of solution:

FIrst observe that $Z \in \mathbb{R}$

If $Z\leq 0$ you can integrate the following region

$$\mathbb{P}[Z>z]=\int_0^{\infty}\lambda e^{-\lambda x}\Bigg[\int_0^{2x-z}\lambda e^{-\lambda y} dy \Bigg]dx=1-\frac{1}{3}e^{\lambda z}$$

Thus $F_Z(z)=\frac{1}{3}e^{\lambda z}$

If $Z>0$ you integrate the following region

$$\mathbb{P}[Z>z]=\int_z^{\infty}\lambda e^{-\lambda x}\Bigg[\int_0^{2x-z}\lambda e^{-\lambda y} dy \Bigg]dx=e^{-\lambda z}-\frac{1}{3}e^{-2\lambda z}$$

Putting the result together you find the CDF of Z

$$ F_Z(z) = \begin{cases} \frac{1}{3}e^{\lambda z}, & z\leq 0 \\ 1-e^{-\lambda z}+\frac{1}{3}e^{-2\lambda z}, & z>0 \end{cases}$$

This is the drawing of $F_Z(z)$ for a fixed $\lambda=1$

As you can see, this is a nice CDF.

To find the desired pdf, just take the derivative of F