Is an example of a PDF $\lambda e^{-\lambda t}$ or $\lambda e^{-\lambda t}\mathrm{d}t$?

Question

The reason I ask is because according to this source:

the $\fbox{$\color{blue}{\mathrm{PDF}}$}$ for the sum of two Exponential Density Functions is

$$\rho(x_1,x_2)\mathrm{d}x_1 \mathrm{d}x_2=\lambda e^{-\lambda x_1}\lambda e^{-\lambda x_2}\mathrm{d}x_1 \mathrm{d}x_2=\lambda^2 e^{-\lambda(x_1+x_2)}\mathrm{d}x_1 \mathrm{d}x_2$$

So this implies that a PDF must be written in the form $$\rho(z)\mathrm{d}z$$ but this makes no sense as I thought that $\rho(z)$ is the PDF.

Now part c) of this question and solution says that:

the combined $\fbox{$\color{blue}{\mathrm{PDF}}$}$ for independent events is $$\lambda e^{-\lambda x_1}\lambda e^{-\lambda x_2}=\lambda^2 e^{-\lambda(x_1+x_2)}$$

This is clearly a contradiction as they can't both be correct. So which one is correct and why?

Thanks.

Answer 1

The initial sentence on the linked page is wrong. The author is being sloppy with language. The pdf in the first line is the joint pdf of the two random variables, not the pdf of the sum. And it isn't even quite that. The function $$ \lambda^2 e^{-\lambda(x_1+x_2)} \quad\text{for }x_1,x_2>0 $$ is the probability density with respect to the measure $dx_1\,dx_2$. One can then say that $$ \lambda^2 e^{-\lambda(x_1+x_2)}\, dx_1 \, dx_2 $$ is the probability measure or the probability distribution (not to be confused with the cumulative distribution function, often called simply the "distribution function").

Answer 2

The exponential distribution is part of the gamma distribution family, if $X$ has ${\Gamma(\alpha,\beta)}$ distribution: $$ f_X(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} $$ $\alpha$ is the shape parameter (more important) and $\beta$ is the scale parameter. The exponential distribution with parameter $\lambda$ is $\Gamma(1,\lambda)$, and it is easy to see, that the sum of $k$ exponentials has distribution $\Gamma(k,\lambda)$. So the first answer is incorrect, it should be $$ f_{X_1+X_2}(y)=\lambda ^2 y e^{-\lambda y} $$ They meant that $$ f_{X_1+X_2}(y)=f_{X_1}*f_{X_2}(y)=\int_{-\infty}^\infty f_{X_1}(t)f_{X_2}(y-t)\mathrm{d}t=\int_{0}^y\lambda e^{-\lambda y}\lambda e^{-\lambda (t-y)}\mathrm{d}y $$