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Let's $T_1$ be a random variable with pdf:

$$f(t) = \frac{6+2t}{7}$$

and $T_2 \sim Exp(\frac{1}{3})$

Knowing that $T_1$ and $T_2$ are independent calculate

$$P(T_1 + T_2 > 1) $$

During my classes we produced the following result:

$$\int_0^1 \int_{1-t}^{\infty} \frac{6+2t}{7} \frac{1}{3} e^{-\frac{1}{3}x} dxdt $$

Could somebody explain me how it was obtained (espacially the limits of integragtion)?

Thanks in advance

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Are you sure you've correctly stated that random variable's pdf? Is it perhaps a piecewise function, instead? Also, is there the assumption that $T_1$ and $T_2$ are independent (as it seems must be the case)? – Cameron Buie Nov 8 '13 at 20:42
Yes, denominator should be $7$. Sorry for a problem. In my notes $t$ was pretty similar to $7$. And, there is an assumtption of $T_1$ and $T_2$ independence – JosephConrad Nov 8 '13 at 20:47
up vote 3 down vote accepted

I suspect that the pdf of $T_1$ is supposed to be $$f(t_1)=\begin{cases}\cfrac{6+2t_1}7 & \text{for }0\le t\le1\\0 & \text{otherwise},\end{cases}$$ since without that requirement, it is not a probability distribution, and the limits of integration later make no sense.

Since $T_1,T_2$ are independent, then their joint pdf is simply the product of their individual pdfs, namely: $$g(t_1,t_2)=\begin{cases}\cfrac{6+2t_1}7\cdot\cfrac13e^{-\frac13t_2} & \text{for }0\le t\le1,t_2>0\\0 & \text{otherwise}.\end{cases}$$ Now, observing that $T_1+T_2>1$ if and only if $T_2>1-T_1,$ we have $$P(T_1+T_2>1) = \int_{-\infty}^\infty\int_{1-t_1}^\infty g(t_1,t_2)\,dt_2\,dt_1.$$ Since $g(t_1,t_2)=0$ for $t_1<0$ and for $t_1>1,$ then $$P(T_1+T_2>1) = \int_0^1\int_{1-t_1}^\infty g(t_1,t_2)\,dt_2\,dt_1.$$ Since $1-T_1\ge 0$ when $T_1\le 1,$ then $T_2>0$ when $T_2>1-T_1$ and $T\le 1,$ and so $$P(T_1+T_2>1) = \int_0^1\int_{1-t_1}^\infty \cfrac{6+2t_1}7\cdot\cfrac13e^{-\frac13t_2}\,dt_2\,dt_1.$$

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Ok, thanks for help:) Really great explanation!! – JosephConrad Nov 8 '13 at 21:22

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