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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...

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Is it the probability density you mean? – Golbez Oct 18 '12 at 16:05
Yeah..Thats what I mean – shaunshd Oct 18 '12 at 16:09
That's not a probability density function, its integral is not $1$. Probably what is intended is $\frac{1}{2\sqrt{3}}$ when $|s_1|\le \sqrt{3}$. – André Nicolas Oct 18 '12 at 16:34
up vote 0 down vote accepted

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 $\dfrac{1}{12}$.

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Thank you...... – shaunshd Jan 3 '13 at 3:33

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