# sum of two independent exponential distributions?

The sum of two independent exponential distributions \begin{aligned}{Z=\omega_{1}X_{1}+\omega_{2}X_{2}}\end{aligned} is as follows. \begin{aligned}f_{Z}(z)=&\int _{0}^{z}\omega_{1}\omega_{2}f_{X_{1}}(x_{1})f_{X_{2}}(z-x_{1})\,dx_{1}\\=\omega_{1}\omega_{2}&\lambda _{1}\lambda _{2}\exp[-\lambda _{2}z]\int _{0}^{z}\exp[(\lambda _{2}-\lambda _{1})x_{1}]\,dx_{1}\\=&{\frac {\omega_{1}\omega_{2}\lambda _{1}\lambda _{2}}{\lambda _{2}-\lambda _{1}}}\left(\exp[-\lambda _{1}z]-\exp[-\lambda _{2}z]\right)\end{aligned}

But the following example doesn't follow the above equation. \begin{aligned}\omega_1=exp(-0.114989504)\end{aligned} \begin{aligned}\omega_2=exp(-7.5448246293224)\end{aligned} \begin{aligned}\lambda_1=1/(4.3258*10^{-08})\end{aligned} \begin{aligned}\lambda_1=1/(1.8737*10^{-03})\end{aligned}

The result is compared with conv(fX1,fX2)*(x_interval) in Matlab.

Can you tell me what the problem is?

• The convolution in Matlab is a summation, not an integral. My guess is that you reached limits of numerical integration. – P. Quinton Mar 23 at 12:47
• I reduced the x_interval for the convolution. And I solve that problem. Thank you so much. – kwonhj Mar 23 at 12:55
• It's not really about the interval, it's more about the fact that sometimes $\sum_n f(n\Delta) \Delta$ may not converge to $\int f(x) dx$ when $\Delta$ goes to $0$ – P. Quinton Mar 23 at 12:58