# Triple Integral limits help

I have 4 random variables: $p_1, p_2, p_3, p_4$

The joint probability distribution function of $p_1, p_2, p_3$ is:

$f(p_1,p_2,p_3) = p_1^{b_1 + x_1 - 1} p_2^{b_2 + x_2 - 1} p_3^{b_3 + x_3 - 1} (1-p_1-p_2-p_3)^{x4+b4-1}$

where $x_1,x_2,x_3,x_4,b_1,b_2,b_3,b_4$ are positive real numbers.

I want to triple integrate f($p_1, p_2, p_3$) over the following criteria:

$p_1 + p_2 + p_3 + p_4 = 1$

$p_1 + p_2 + p_3 \leq 1$

$p_1 > p_2$

$p_1 > p_3$

$p_1 \geq 0$

$p_2 \geq 0$

$p_3 \geq 0$

$p_4 \geq 0$

Can someone tell me what should be the limits of my triple integral and in which order f($p_1, p_2, p_3$) should be integrated?

First we select one variable, let's say $p_1$. It can be any number greater or equal $0$. Also it have to fulfill the condition $p_1+p_2+p_3\leq 1$. So far we know nothing about $p_3$ and $p_2$, only that these are nonnegative. So we have:

$p_1\leq 1-p_2-p_3 \leq 1-0-0 = 1$

Thus $p_1 \in [0, 1]$.

Then we select the second one, eg. $p_2$. We know, that it is nonnegative, smaller than $p_1$ and also if we add it to $p_1$ and $p_3$ it must be smaller or equal $1$:

$p_1+p_2 +p_3 \leq 1 \\ p_2 \leq 1-p_1-p_3$

So far we doesn't have selected $p_3$, but we know, that $p_3$ is nonnegative ($p_3\geq 0$), so the inequality in its' broadest way will take the form

$p_2 \leq 1-p_1-p_3\leq 1-p_1-0 = 1-p_1$

So $p_2 \in [0, \min(p_1,1-p_1)]$.

And now we can select $p_3$ - nonnegative, smaller than $p_1$ and in sum with $p_1$ and $p_2$ smaller or equal than $1$. So $p_3 \in [0, \min(p_1, 1-p_1-p_2)]$.

Now we can describe our area as following: $$V=\begin{cases}p_1 \in [0,1]\\ p_2 \in [0, \min(p_1,1-p_1)]\\ p_3 \in [0, \min(p_1, 1-p_1-p_2)]\end{cases}$$

The upper boundary for $p_2$ changes in $p_1=.5$, for $p_3$ in $p_1=\frac{1}{3}$, so we can write this area as disjoint sum of the following three areas: $$V_1=\begin{cases}p_1 \in [0,\frac{1}{3}]\\ p_2 \in [0, p_1]\\ p_3 \in [0, p_1]\end{cases}$$ $$V_2=\begin{cases}p_1 \in [\frac{1}{3}, .5]\\ p_2 \in [0, p_1]\\ p_3 \in [0, 1-p_1-p_2]\end{cases}$$ $$V_3=\begin{cases}p_1 \in [.5,1]\\ p_2 \in [0, 1-p_1]\\ p_3 \in [0, 1-p_1-p_2]\end{cases}$$

The main idea used in this approach is to describe boundaries using only variables selected before - so $p_1$ is independent, $p_2$ depends only on $p_1$ and $p_3$ depends on $p_1$ and $p_2$.

Integration order is from the most dependent variable to the independent one, so in this case you first integrate by $dp_3$, then by $dp_2$ and then by $dp_1$

You can also try another orders of selecting variables. If the conditions are 'symmetric', like conditions for $p_2$ and $p_3$, you should obtain 'symmetric' results, eg.: $$V=\begin{cases}p_1 \in [0,1]\\ p_3 \in [0, \min(p_1,1-p_1)]\\ p_2 \in [0, \min(p_1, 1-p_1-p_3)]\end{cases}$$

Simplier case

It sems, that there is a problem with understanding the boundaries built on the condition $p_1+p_2+p_3\leq 1$, so let's try explaining it in other way:

Let's say that you have 1 liter bucket of icecream and three cupd to fill them with icecream. You don't have to spend all the icecream and you can leave some cups empty. The conditions for this problem are:

$x\geq 0; y\geq0; z\geq 0; x+y+z \leq 1;$

What you do?

1. First you take $x\in [0,1]$ litres of icecream and put it into the first cup (notice, that there are no $y$ and $z$ in these boundaries.
2. Then you take $y\in [0,1-x]$ litres of icecream and put it into the second cup (ammount of icecream you can take is limited only by $x$, that is already taken!)
3. Then you take $z\in [0,1-x-y]$ litres of icecream and put it into the third cup

Do you get it?

• Can you explain how you got this? And according to this, the integral will be split in 3 parts, and the order of integration should be $dp_3$ followed by $dp_2$ followed by $dp_1$? – ProgSnob Jun 19 '17 at 8:11
• Edited the original post a bit. $p_1$, $p_2$, $p_3$, $p_4$ are all greater than or equal to 0. – ProgSnob Jun 19 '17 at 8:30
• Shouldn't $p_2 \in [0, min(p_1, 1 - p_1 - p_3]$? After all, the limits should tell be symmetric, and $p_2$ and $p_3$ should be interchangable. – ProgSnob Jun 19 '17 at 8:52
• @ProgSnob I've edited my answer. – Jaroslaw Matlak Jun 19 '17 at 10:06
• Why do you say that $p_2$ is only dependent on $p_1$ when it is in the equality $p_1 + p_2 + p_3 \leq 1$ too. We have $p_1 > p_2$ and $p_1 > p_3$. $p_2$ and $p_3$ should be interchangeable. Also, in the start you write: $p_1 + p_2 + p_3 \leq p_1 + p_2$. But this would imply $p_3 \leq 0$. This seems incorrect. – ProgSnob Jun 19 '17 at 10:38