aide da go podkarame bavno i podred, t.e.
purvo,
f(x) = 2/A - 2x/A, is the pdf of X on [0,A]
posle,
g(y) = 2/B - 2y/B, is the pdf of Y on [0,B]
(t.e. naklona im vurvi ot liavo-nadiasno-nadolu)
i tuikato sa independent, to
joint pdf of (X,Y) is the product f(x)g(y)
and, the support of the two-dimensional random
variable (X,Y) is the rectangle [0,A]times[0,B]

a, sega se borim za distribution-a na Z = X+Y,
mojem da namerim naprimer
cdf of Z, and we're done,
i.e. P(Z < z) = P(X+Y < z) = integral of f(x)g(y)
over the support, i.e., over the rectangle [0,A]times[0,B],
and now, there you go, everything depends on z,
everything = limits of integration, that is, we
have to consider 3 cases:
z < B; B < z < A; and A < z < A+B,
hopefully, you can now figure out the x-and-y
limits of integration ...
this is surely the hardest approach ;-)
(btw, pravi razlika mejdu Z i z ...)
another one is the change of variable technique,
(dvumerna smiana!)
for example, Z=X+Y and X=X, etc.
you finally get the pdf of Z,
and,
the third aproach could be the convolution, i.e.
the pdf of Z is the convolution of X and Y,
sirech
neka h(z) e toia pdf, togava
h(z) = integral of f(x)g(z-x)
... i pak vsichko zavisi ot "z" ...
t.e. podintegralnata funkcia e edna za z<B,
i druga za B<z<A, a suvsem treta za A<z<A+B,
koeto vodi i do razlichni granitci za x pri integriraneto po x,
t.e. niakoi put / v niakoi sluchai integrala e
ot 0 do z, no drug put e ot z-B do z, etc.
hint: vij definiciata na g(.)
... moje i po slojno da se zamotae ;;;)))
... ne mu viarvai mnoo mnoo