I have the following problem:
$Z = 3 * RND_1 - 5* RND_2 $, where $RND_1 $ and $RND_2$ are two independent uniformly distributed random variables they take values from (0,1) and we compose an equation of them this gives us Z
What is the $F(Z)$ - cumulative distribution function (CDF)?
I understand for solving this kind of problems i need to draw lines and then calculate the area where the inequality holds. When my teacher was explaining he confused himself and couldn't make the problem clear
For simplier notation let say $RND_1 = x ,$ $RND_2 = Y$
$Z = 3X - 5Y$
so $F(z) = p(Z < z) = p(3X - 5Y < z)$ ==
$3X - 5Y < z$ can be written as $\frac{3X}{5}-\frac{z}{5} < 5$
== $p(\frac{3X}{5}-\frac{z}{5} < Y$)
As i am writting this question i am starting to understand better the problem. I need to determine where is the inequality holds and then subtract this area from the whole square's.
Can someone make the whole thing clear?
off topic:Please if you have any probability exercise book recommendation just let me know the name of the book.
$Z = 3 X - 5Y$ so $P(Z < x) = P(3 X - 5 Y < x) = ?$ and Then you write this probability as an integral (try to justify for yourself why it holds) $$\int_{xx=0}^1 \int_{yy=0}^1 \delta(3x - 5 y < x) dxx dyy$$ where $\delta(3x - 5 y < x)$ is equal to 0 is the inequality is not respected. And then you can see that this integral is only making an integration on a part of a square.