# Does my claim work?

Suppose $X\sim\mathrm{Exp}$ and $Y$ is a non negative independent random variable. Does $P(X-Y>t\mid X>Y)=P(X>t)$, where $t$ is a non negative real number. I think it is true as it seems to be can treat the random variable $Y$ as some constants but not sure whether it is correct.

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Observe: $$P(X-Y > t|X>Y) = \int_{0}^{\infty}P(X>y+t|X>y,Y=y)dF_Y(y)$$ $$= \int_{0}^{\infty}P(X>y+t|X>y)dF_Y(y)$$ due to independence $$= \int_{0}^{\infty}P(X>t)dF_Y(y)$$ due to memoryless property of exponential $$= P(X>t)\int_{0}^{\infty}dF_Y(y) = P(X>t)$$
your first integral should integrate with respect to $y$ instead of $F_Y(y)$, aren't it? –  Mathematics Dec 21 '12 at 5:24
If Y has a pdf say $f_y(y)$ then replace $dF_y(y)$ with $f_y(y)dy$. If not, then we have to use the Riemann Stieltjes form which is what I have written. –  Gautam Shenoy Dec 21 '12 at 12:47