Given that $T_1, T_2$ are iid $\text{exp}(\lambda)$ variates.

I want to find the cdf $F_T(t)$ where $T=T_1-T_2$

My Attempt

$F_T(t) =_1 P(T<t) = P(T_1-T_2<t) = P(T_1<T_2 + t)$

Where $=_1$ is true according to the definition of CDF.

But I'm stuck here, how can I continue from this step?


2 Answers 2


Assuming independence between $T_1$ and $T_2$ you have to evaluate the following integral

$$F_T(t)=\int \int_{T<t}f_{T_1}(t_1)f_{T_2}(t_2)d t_1 d t_2$$

thus do a drawing of the integration region and solve the double integral.

Observe that integral bounds change according with $T>0$ or $T<0$


If $T_{1}$ and $T_{2}$ are independent then

If $t\geq 0$,

$$P(T\leq t)=P(T_{1}-T_{2}\leq t)=\mathbb{E}(\mathbf{1}_{\{T_{1}-T_{2}\leq t\}})\\=\mathbb{E}(\mathbb{E}(\mathbf{1}_{\{T_{1}\leq t-t_{2}\}}|T_{2}=t_{2}))=\int_{0}^{\infty}\int_{0}^{t+t_{2}}\lambda^{2}e^{-\lambda (t_{1}+t_{2})}dt_{1}\,dt_{2}=1-\frac{e^{-\lambda t}}{2}$$.

This evaluates to

If $t<0$. Then

$$P(T\leq t)=\mathbb{E}(\mathbb{E}(\mathbf{1}_{\{T_{1}\leq t-t_{2}\}}|T_{2}=t_{2}))=\\\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\{t_{2}+t\geq 0,t_{1}\leq t_{2}+t\}}\lambda^{2}e^{-\lambda (t_{1}+t_{2})}dt_{1}\,dt_{2}=\\\int_{-t}^{\infty}\int_{0}^{t+t_{2}}\lambda^{2}e^{-\lambda (t_{1}+t_{2})}dt_{1}\,dt_{2}=\frac{1}{2}\cdot e^{\lambda t}$$.

So :-

$$\large F_{T}(t)=\begin{cases} 1-\frac{e^{-\lambda t}}{2}\quad, t\geq 0\\\frac{e^{\lambda t}}{2}\quad,t<0\end{cases}$$

The pdf can be found correspondingly by differentiating wrt $t$.

$$\large f_{T}(t)=\begin{cases} \frac{\lambda e^{-\lambda t}}{2}\quad, t\geq 0\\\frac{\lambda e^{\lambda t}}{2}\quad,t<0\end{cases}$$

  • $\begingroup$ All of them are exponential integrals. I don't think you will have trouble in evaluating them. $\endgroup$ Dec 22, 2021 at 8:58
  • $\begingroup$ @dan I have edited my answer. $\endgroup$ Dec 22, 2021 at 9:12
  • $\begingroup$ What are those E's? Expected value maybe? Can't the above be re-written without using E in the formula (I didn't learn what are those ones to...) $\endgroup$
    – Dan
    Dec 22, 2021 at 9:23
  • $\begingroup$ I think you have something wrong with integral limits, I don't understand why u split to cases when t<0 and when t>=0... $\endgroup$
    – Dan
    Dec 22, 2021 at 9:33
  • $\begingroup$ $\mathbb{E}$ stands for expected value. And you can write the integrals in terms of the joint pdf. The limits will be the same(in fact what I did is the justification of why that is the case). If you are having trouble understanding why I split those integrals then you should brush up the definitions as I don't think your concepts are clear. $\endgroup$ Dec 22, 2021 at 9:47

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