probability density and distribution

Question

Let $X$ and $Y$ be uniform random variables where $0\leqslant x,y\leqslant 1$.

Let $\operatorname{sgn}(x)$ be $1$ when $x>0$, $−1$ when $x<0$ and $0$ when $x=0$. Find the distribution and density of $\operatorname{sgn}(x-1/2)+\operatorname{sgn}(y-1/2)$.

I have calculated the density of $\operatorname{sgn}(x-1/2)=\operatorname{sgn}(y-1/2)= f(z)=1/2$ when $z=1$ and $1/2$ when $z=-1$.

I feel $\operatorname{sgn}(x-1/2)+\operatorname{sgn}(y-1/2)=0$ but I'm not sure.

Any help or ideas would be appreciated. Thank you!

what is $y$? Do you mean $X,Y$ are iid and find the distribution of $sgn(X-1/2)+sgn(Y-1/2)$? — Dima McGreen, Aug 28 '13 at 10:50
Do you mean to say that $X$ and $Y$ are uniformly distributed? If so, you need to specify two parameters. Else, please explain what you mean to say. — dreamer, Aug 28 '13 at 11:05
the parameters are 0<=x<=1 and 0<=y<=1 in the interval [0,1] — user91689, Aug 28 '13 at 11:22
In the question we are given that a point within the coordinate (X,Y) is choosen uniformly at random from a square:(x,y) belongs to R^2: 0<=x<=1 and 0<=y<=1 — user91689, Aug 28 '13 at 11:27

score 1 · Answer 1 · edited Aug 28 '13 at 12:56

Let $z$ be your new variable $z=\operatorname{sgn}\left(x-\frac{1}{2}\right) +\operatorname{sgn}\left(y-\frac{1}{2}\right)$. As sign function is discrete, so it will be $z$. Concretely, it will have three possible values:

$z=-2$ in case both $x$ and $y$ are below $\frac{1}{2}$
$z=2$ in case both $x$ and $y$ are above $\frac{1}{2}$
$z=0$ otherwise

Taking into account that $x$ and $y$ are independent, one can find the density function of $z$ as: Prob$(z=-2)=0.25$, Prob$(z=0)=0.5$, and Prob$(z=2)=0.25$.

probability density and distribution

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