Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
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

1 Answer 1

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 $\dfrac{1}{12}$.

share|improve this answer
    
Thank you...... –  shaunshd Jan 3 '13 at 3:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.