Is my evaluation for the cumulative distribution function (CDF) right?

30 Views Asked by At

I want to evaluate the CDF of $X+Y-\ln(Y)$ where $X$ and $Y$ are independent random variables that can only have non-negative values. To find $P(X+Y-\ln(Y)<\tau)$ I have $$P(X+Y-\ln(Y)<\tau)=\int_{l_1}^{l_2}F_X(\tau-y+\ln(y))f_Y(y)dy~~~~\text{Eq. }1$$ where $F_X(x)$ is the CDF of $X$, $f_Y(y)$ is the density function of $Y$ and $l_1$ and $l_2$ are two roots of the function $\tau-y+\ln(y)$ with $l_1 \leq l_2$. Eq. 1 can also be written as $$P(X+Y-\ln(Y)<\tau)=\int_{l_1}^{l_2}\int_0^{\tau-y+\ln(y)}f_X(x)f_Y(y)dxdy$$ Now using $y=v$ and $u=x+v$ we can write $$P(X+Y-\ln(Y)<\tau)=\int_{l_1}^{l_2}\int_v^{\tau+\ln(v)}f_X(u-v)f_Y(v)dudv.~~~~~~\text{Eq. }2$$ Is my Eq. 2 right? Or should it be $$P(X+Y-\ln(Y)<\tau)=\int_{l_1}^{l_2}\int_{e^{u-\tau}}^{u}f_X(u-v)f_Y(v)dvdu. ~~~~\text{Eq. }3$$ I obtained Eq. 3 from a research paper. I do not know how Eq. 3 is correct. Please explain why equation 3 is correct. Thanks in advance.