<s>[INST] Probability that Modulus of Difference Greater than $3$ Given that: $X$ and $Y$ be two Continuous Random Variables with Joint pdf
$$f_{XY}(x,y)=\frac{1}{x^2y^2}, x \ge 1\,\,,y \ge 1$$ Find $$P\left(|Y-X|\gt 3\right)$$
My Try: Its Clear that $X$ and $Y$ are Independent and Identically Distributed i.e.,
$$f_X(x)=\frac{1}{x^2}, x \ge 1$$ and $$f_Y(y)=\frac{1}{y^2}, y \ge 1$$
Now $$P\left(|Y-X|\gt 3\right)=1-P\left(|Y-X|\le 3\right)=1-P(-3 \le Y-X \le 3)=1-P(X-3 \le Y \le X+3)$$ Now
$$P(X-3 \le Y \le X+3)=\int_{x=4}^{\infty}P((X-3 \le Y \le X+3)|X=x)f_X(x)dx$$ $\implies$
$$P(X-3 \le Y \le X+3)=\int_{x=4}^{\infty}P(x-3 \le Y \le x+3)f_X(x)dx$$ $\implies$
$$P(X-3 \le Y \le X+3)=\int_{x=4}^{\infty}\left(\int_{y=x-3}^{x+3}f_Y(y)dy\right)f_X(x)dx$$
$$\int_{y=x-3}^{x+3}f_Y(y)dy=\frac{6}{x^2-9}$$ $\implies$
$$P(X-3 \le Y \le X+3)=\int_{x=4}^{\infty}\frac{6}{(x^2-9)x^2}=\frac{6}{9}\int_{4}^{\infty}\frac{1}{x^2-9}-\frac{1}{x^2}=\frac{Ln7}{9}-\frac{1}{6}=0.049545$$ So the Required Probability is $1-0.049545=0.95$
Let me know any other approach to this Problem, Thanks.
 [/INST] Compute $P(Y\gt X+3)=p$, where
$$
p=\int_1^\infty\frac{\mathrm dx}x^2\int_{x+3}^\infty\frac{\mathrm dy}{y^2}=\int_1^\infty\frac{\mathrm dx}{x^2(x+3)},
$$
then use the identity
$$
P(|Y-X|\gt3)=P(Y\gt X+3)+P(X\gt Y+3)=2p.
$$
(Note that $p$ cannot involve $\log7$, only $\log4$.)
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