CDF and PDF of $Z = aX +bY$ ; $X \sim \exp({\lambda_1})$ and $Y \sim \exp({\lambda_2})$

Question

I would like to derive the Cumulative Distribution Function and Probability Density Function of a random variable ($Z$) which is a linear combination of two exponential random variables with different parameters. i.e.

$Z = aX +bY$ ; $X \sim \exp({\lambda_1})$ and $X \sim \exp({\lambda_2})$ ; where, $a$ and $b$ are positive constants

$f_X (x) = \lambda_1 e^{\lambda_1 x}$

$f_Y (y) = \lambda_2 e^{\lambda_2 y}$

Please note that $X$ and $Y$ are independent.

I searched through the internet and I found the answer for $Z = X + Y$ is a hypoexponential distribution, but I couldn't find an answer for a linear combination.

Could you please explain the methods and possible some references for me to learn this ?

Answer 1

specifying some details would be appreciated

1. $X$ and $Y$ are independent?

2. $a,b$ are real parameters? or non negative? (this changes a lot the calculation)

For the rest you can use the standard CDF's method

$$F_Z(z)=\int_{aX+bY \leq z}f_{XY}(x,y)dxdy$$

and

$$f_Z(z)=\frac{d}{dz}F$$

Hint for the calculations

1. CDF's Method

$$F_Z(z)=\mathbb{P}[Z \leq z]=\mathbb{P}[aX+bY \leq z]=\mathbb{P}[Y \leq \frac{z}{b}-\frac{a}{b}X]$$

This is the drawing

Then this is the integral to solve

$$F_Z(z)=\int_{0}^{\frac{z}{a}}\lambda e^{-\lambda x}dx\int_{0}^{\frac{z}{b}-\frac{a}{b}x}\theta e^{-\theta y}dy$$

[I changed the exp parameters in $\lambda$ and $\theta$ to simplify the notation]

Other methods:

2. Use the Fundamental Transformation Theorem (Jacobian Method)

3. Considering that if $X\sim exp(\theta)$ then $aX\sim exp(\frac{\theta}{a})$ you can modify the parameters of your marginal distribution and calculate the density of the sum immediately by convolution

The first method is useful to improve your brainstorming

Answer 2

A variation of in @tommik's solution #1, if $X$ and $Y$ are independent rvs, you can directly use the convolution formula for densities, with Lebesgue measure $\lambda$ on $\mathbb{R}$: $$ (f \ast g)(x) = \int_{A}f(A-y)h(y)\lambda(dy) $$ Note that $f_X(x)=0 \ \forall x<0$, so we want $\frac{z-by}{a}>0:f(\frac{z-by} {a})>0$, which implies upper bound of $y<\frac{z}{b}$. \begin{align} (f \ast g)(x) &= h_Z(z) = \int_{0}^{\frac{z}{b}}f_X(\frac{z-by}{a})h_Y(y)dy = \lambda_1 \lambda_2\int_{0}^{\frac{z}{b}}e^{-\lambda_1(\frac{z-by}{a})}e^{-\lambda_2 y}dy\\ &=\lambda_1 \lambda_2e^{-\frac{\lambda_1 z}{a}}\int_{0}^{\frac{z}{b}}e^{-(\lambda_2 - \frac{\lambda_1b}{a})y}dy = \frac{a\lambda_1 \lambda_2 e^{-\frac{\lambda_1 z}{a}}}{a\lambda_2-b\lambda_1 } \bigg(1-e^{-\frac{(a\lambda_2-b\lambda_1)z}{ab}}\bigg)\\ \end{align} Note this solution works only when $\lambda_1 \neq \lambda_2$ and $X,Y$ are independent. Also, either both $a,b>0$ or both $a,b<0$