The given solution :
I find it really counter-intuitive that how can the probability that random variable take the value 1 = 1/4. I thought in continuous random variable, the PDF output the density and that is why we take the area. And how is the joint PDF of x and y when y = 1 is still 1/16.
Doesn't it change once we assign the value 1 to the random variable Y ?
Can someone please explain.
When you fix $Y=1$ then you can't simply use joint PDF (a function of $x$ and $y$) and say that the joint PDF of x and y when y=1 is still 16. Because Y is now known, it's no longer random so it's not a function of $x$ and $y$ any more.
Instead you need to rely on the following definition of $f_{XY}(x, y)$ or just $f(x, y)$:
$$ P\{X\in A,Y\in B\} = \int_{B}^{}\int_{A}^{}f(x,y)dx dy $$
Or equivalently:
$$ f(x, y)= \frac{\partial^2}{\partial x\partial y}F(x,y) $$
Where (suppose $X\in\mathbb{R}, Y\in \mathbb{R}$): $F(x,y)=P\{X\in (-\infty, x], Y\in (-\infty, y] \}$ is the joint cumulative probability distribution of $X$ and $Y$.
So if you substitute $Y=1$ what you get is:
$$ f(x, 1)= \frac{\partial^2}{\partial x\partial y}F(x,y)|_{y=1}, $$
a function of single variable $x$ that is often used to calculate the PDF of $X$ conditioned on $Y$ (as you see in the solution).
No it doesn't because $f(x, y)$ is defined to behave as such, though it may change if, for example, $f(x, y)=2+y$ (of course you should have different bounds for $X$ and $Y$ for it to be a valid joint PDF).