I have a problem where I have two independent variables each having a probability density function given by:
$p(s_1) = \frac{1}{2}\sqrt{3}$, when $s_1\leq\sqrt{3}$
and $0$, otherwise
And the probability density function is the same for other variable.
When a joint probability function is graphed it says that it will be a square. How?
Thanks for any help…
Best Answer
At the time I am writing this, the claimed probability density function is not a pdf, since its integral is not $1$.
Probably what is intended is $\dfrac{1}{2\sqrt{3}}$ when $|s_1|\le \sqrt{3}$.
We will change notation a little, and assume that we have two independent random variables $X$ and $Y$. Random variable $X$ has pdf $\dfrac{1}{2\sqrt{3}}$ when $|x|\le \sqrt{3}$, and $0$ when $|x|\gt \sqrt{3}$. Random variable $Y$ has pdf $\dfrac{1}{2\sqrt{3}}$ when $|y|\le \sqrt{3}$, and $0$ when $|y|\gt \sqrt{3}$.
Since $X$ and $Y$ are independent, their joint pdf is the product of the individual pdf.
Thus the joint pdf $f(x,y)$ is equal to $\dfrac{1}{12}$ when both $|x|$ and $|y|$ are $\le \sqrt{3}$, and $0$ elsewhere.
So the joint pdf is the constant $\dfrac{1}{12}$ on and inside the square with corners $(\sqrt{3},\sqrt{3})$, $(-\sqrt{3},\sqrt{3})$, $(-\sqrt{3},-\sqrt{3})$, and $(\sqrt{3},-\sqrt{3})$.
If we decide to ignore the parts of the world where the joint pdf is $0$, we have a constant density function on a square. A constant density function on a square is not the same thing as a square, but when we graph $z=f(x,y)$ in space, we will get a square "table" of constant height $\dfrac{1}{12}$.