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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?

Mr.Gandalf Sauron
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Dan
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2 Answers2

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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 $T<0$

tommik
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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 $t$.

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

Mr.Gandalf Sauron
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  • All of them are exponential integrals. I don't think you will have trouble in evaluating them. – Mr.Gandalf Sauron Dec 22 '21 at 08:58
  • @dan I have edited my answer. – Mr.Gandalf Sauron Dec 22 '21 at 09:12
  • 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 '21 at 09:23
  • 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 '21 at 09:33
  • $\mathbb{E}$ 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. – Mr.Gandalf Sauron Dec 22 '21 at 09:47
  • The case for $t\leq 0$ has to be separate because in that situation when you are integrating wrt to $t_{1}$ , the pdf of $T_{1}$ is zero in the region then $t_{1}\leq 0$. So you have to integrate in the region when $0\leq t_{1}\leq t+t_{2}$ and for this to happen , $t_{2}$ must be such that $t+t_{2}\geq 0$ . – Mr.Gandalf Sauron Dec 22 '21 at 09:49
  • This predicament is no longer there when we assume $t\geq 0$. Because in that case $t+t_{2}$ is always non-negative and you can integrate as I did in the first case – Mr.Gandalf Sauron Dec 22 '21 at 09:50
  • Also @Dan if you use the pdf of $T_{1}-T_{2}$ as given in the comments https://math.stackexchange.com/questions/115022/pdf-of-the-difference-of-two-exponentially-distributed-random-variables . You will end up with the same result as I did . If $x>0$ you integrate from $0$ to $x$ . And if $x<0$ you integrate from $-\infty to x$. Also you can use my cdf and differentiate to get the pdf as given here. – Mr.Gandalf Sauron Dec 22 '21 at 10:08