I have two independent random variables $X,Y$ such that $X\sim\operatorname{Erl}(n,\lambda)$ and $Y\sim\operatorname{Erl}(m,\mu)$. I am trying to find the probability $P(X<Y)$.
I started with the following \begin{align} P(X<Y)&=\int_{0}^{\infty}P(X<Y|X=x)P(X=x)dx\\ &=\int_{0}^{\infty}P(Y>x)P(X=x)dx \end{align}
After simplification, I am stuck and unable to get a closed form for the above integral. But I got a hint that a recursive form exists: if $P(X<Y)=G(n,m)$ then $$(\lambda+\mu)G(n,m)=\lambda G(n-1,m)+ \mu G(n,m-1)$$ I tried to evaluate the above integral by parts, considering the integrand as a product of two functions dependent on $n$ and $m$. But the problem still persists. Now I wondered: can I proceed to show the recurrence relation by taking $$X=X_1+X_2+ \dots +X_n$$ $$Y=Y_1+Y_2+ \dots +Y_m$$ where $X_i\sim\exp(\lambda)$ and $Y_j\sim\exp(\mu)$? Am I approaching correctly or is there something I am missing?
Perhaps something along the following lines (not checked):
Starting with $\displaystyle P(X<Y)=\int_{0}^{\infty}P(Y\ge x\mid X=x)\,p(x)\,dx=\int_{0}^{\infty}P(Y\ge x)\,p(x)\,dx$ since you have independence
you know $p(x) = \dfrac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!\,}$ and $\displaystyle P(Y\ge x) = \sum_{k=0}^{m-1}\frac{1}{k!}e^{-\mu x}(\mu x)^{k}$
so $\displaystyle P(X<Y)=\sum_{k=0}^{m-1} \dfrac{\lambda^n \mu^k}{(n-1)!\,k!}\int_{0}^{\infty}e^{-(\lambda+\mu) x}x^{k+n-1}\,dx=\sum_{k=0}^{m-1} {k+n-1 \choose n-1}\dfrac{\lambda^n \mu^k }{(\lambda+\mu)^{k+n}}$ which may have a further simplification