What is P(Y1−Y2>3) of a given joint density function?

Question

Let $Y_1 $and $Y_2$ have joint density function:

$$f (y_1, y_2) = e^{-(y_1+y_2)}, \text{for all } y_1 >0,y_2 >0 $$

What is $P(Y_1−Y_2>3)$?

My attempt:

$$P(Y_1−Y_2>3) = P(Y_1−3> Y_2)$$

$$\int_{0}^{\infty}\int_{0}^{y_1-3} e^{-y_1-y_2} dy_2dy_1$$

Evaluating the integral, the value returns: $1-\frac{e^3}{2}$

The correct answer is however as follows:

$$P(Y_1−Y_2>3) = P(Y_1> Y_2+3)$$

$$\int_{0}^{\infty}\int_{y_2+3}^{\infty} e^{-y_1-y_2} dy_1dy_2$$

Which evaluates to: $\frac{1}{2e^3}$

The thing that confuses me, is that we know $P(Y_1−3>Y_2) = P(Y_1> Y_2+3)$. Why does the two integrations evaluate to different values? Have I done something wrong? Or is there a particular rule that I am violating?

Answer 1

The second integral is correct. The first is not and it must be $$\int_{\color{red}{3}}^{\infty}\int_{0}^{y_1-3} e^{-y_1-y_2} dy_2dy_1$$ since $y_2>0$. we have that $0<y_2<y_1-3$ which implies that $0<y_1-3$ or that $3<y_1$ in the outer integral.