# How to Find CDF of $T_1-T_2$? [duplicate]

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

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

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

• Are $T_{1}$ and $T_{2}$ independent? Dec 22, 2021 at 8:34
• @Mr.GandalfSauron yes they are.
– Dan
Dec 22, 2021 at 8:48
• math.stackexchange.com/q/115022/321264 Dec 22, 2021 at 9:50

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

$$F_T(t)=\int \int_{T

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 $$t$$.

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

• All of them are exponential integrals. I don't think you will have trouble in evaluating them. Dec 22, 2021 at 8:58
• @dan I have edited my answer. Dec 22, 2021 at 9:12
• 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...)
– Dan
Dec 22, 2021 at 9:23
• 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...
– Dan
Dec 22, 2021 at 9:33
• $\mathbb{E}$ 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. Dec 22, 2021 at 9:47