Assume random variable $N$ with a pdf of $f\left(N\right)$ and CDF of $F\left(N\right)$. Accordingly, the other two random variables $X$ and $Y$ are defined based on $N$ so that $X=\alpha N$ and $Y=\left(1-\alpha \right)N$, where $0<\alpha <1$. I need to find the joint probability function of $X$ and $Y,$ knowing that they are not independent. So, I tried the following calculations. Is it correct?
$$\begin{align}F_{X,Y}\left(s,t\right) &=\Pr\left(X\le s,Y\le t\right)\\&=\Pr\left(X\le s\right)\cdot \Pr\left(Y\le t\mid X\le s\right)\\&=\Pr\left(\alpha N\le s\right)\cdot \Pr\left(\left(1-\alpha \right)N\le t\mid \alpha N\le s\right)\\&=\Pr\left(N\le \frac{s}{\alpha }\right)\cdot \Pr\left(\left.N\le \frac{t}{1-\alpha }\right|N\le \frac{s}{\alpha }\right)\end{align}$$
If $\dfrac{s}{\alpha }\le \dfrac{t}{1-\alpha }$, $\Pr\left(\left.N\le \dfrac{t}{1-\alpha }\right|N\le \dfrac{s}{\alpha }\right)=1$ and consequently $F_{X,Y}\left(s,t\right)=\Pr\left(N\le \dfrac{s}{\alpha }\right)=F_X\left(\dfrac{s}{\alpha }\right)$
Otherwise, $F_{X,Y}\left(s,t\right)=F_X\left(\dfrac{s}{\alpha }\right).\dfrac{F_Y\left(\frac{t}{1-\alpha }\right)}{F_Y\left(\frac{s}{\alpha }\right)}$.
Please advise.