# Uniform distribution density function

Let's say we have two random variables $T_1$ and $T_2$ and the joint density function of the two is uniform over the region $0\leq t_1\leq t_2 \leq L$, where L is a positive constant. Then the area of the region would equal $\frac{L^2}{2}$.

Now, I need to find the expected value of the sum of the squares, $E[T_1^2+T_2^2]$, which could easily be found by finding the density function, $f(t_1,t_2)$, and computing a double integral of $(t_1^2+t_2^2)\cdot f(t_1,t_2)$ while setting the bounds over the region of the triangle $0\leq t_1\leq t_2 \leq L$.

I'm told the density function here is $\frac{2}{L^2}$. Is the density function of 2 uniformly distributed random variables $\frac{1}{Area}$, just as the density function of 1 uniformly distributed random variable is $\frac{1}{Distance}$?

If so, when dealing with 3 variables will $f(T_1,T_2,T_3) = \frac{1}{Volume}$?

Is there a general rule for an $n$ amount of variables?

• Indeed, if the $n$tuple $(T_1,\ldots,T_n)$ is uniformly distributed on a domain $D$ of $\mathbb R^n$ with volume $|D|$ the density is $1/|D|$ on $D$ and $0$ on $\mathbb R^n\setminus D$, which can be written as $$f_{T_1,\ldots,T_n}=\frac{\mathbf 1_D}{|D|}.$$ – Did Jan 13 '16 at 17:26
• Why the $1_D$ and not just 1? – Itiel Cohen Jan 13 '16 at 17:50
• $\frac1{|D|}$ would denote a positive constant function. Such functions do not work as PDF. Integration of a PDF gives $1$, not $+\infty$ as a positive constant function does. – drhab Jan 13 '16 at 17:57
• Got it, thank you guys. – Itiel Cohen Jan 13 '16 at 18:00