Joint Probability Density Function: f(x,y,z,t) = 1/(xyz) for 0<t<=z<=y<=x<=1, 0 elsewhere

[mario99]

(a) Prove that the following is a joint probability density function.

$\displaystyle f(x,y,z,t) = \begin{cases} \frac{1}{xyz} & \ \ \ ,0<t\leq z\leq y\leq x\leq 1 \\ 0 & \ \ \ , \text{elsewhere} \end{cases}$

(b) Suppose that $f$ is the joint probability density function of random variables $X, Y, Z, \text{and} \ T$. Find $f_{Y,Z,T}(y,z,t), f_{X,T(x,t)}, \text{and} \ f_{Z(z)}.$


My attempt.

In part (a), it wants me to solve the integral of the function $f(x,y,z,t)$. The answer must equal to $1$.

$\displaystyle \int\int\int\int \frac{1}{xyz} \ dx \ dy \ dz \ dt$

The domain is confusing. How to sit the limits of integration?

In part (b), I am totally lost.

 

[BigBeachBanana]

For a) state the domain more explicitly.

• For t, the limits are $0 < t < 1.$
• For z, the limits are $t \leq z \leq 1.$
• For y, the limits are $z \leq y \leq 1.$
• For x, the limits are $y \leq x \leq 1.$

It follows

$$\int_0^1 \int_t^1 \int_z^1 \int_y^1 \dfrac{1}{xyz} \, dx\, dy \, dz \, dt$$
For b) once find a) you can find the marginal density function by integrating with respect to the variable of interest while keeping the others constant.

 

[mario99]

Thank you BigBeachBanana


$\displaystyle \int_0^1 \int_t^1 \int_z^1 \int_y^1 \dfrac{1}{xyz} \, dx\, dy \, dz \, dt = \displaystyle \int_0^1 \int_t^1 \int_z^1 \dfrac{-\ln y}{yz} \, dy \, dz \, dt$


$\displaystyle = \displaystyle \int_0^1 \int_t^1 \dfrac{\ln^2 z}{z} \, dz \, dt = \displaystyle \int_0^1 -\ln^3 t \, dt = 6 \neq 1$


What was wrong?

 

[BigBeachBanana]

$$\displaystyle \int_0^1 \int_t^1 \red{\dfrac{\ln^2 z}{z}} \, dz \, dt$$
This is wrong.

 

[khansaheb]

Thank you BigBeachBanana


$\displaystyle \int_0^1 \int_t^1 \int_z^1 \int_y^1 \dfrac{1}{xyz} \, dx\, dy \, dz \, dt = \displaystyle \int_0^1 \int_t^1 \int_z^1 \dfrac{-\ln y}{yz} \, dy \, dz \, dt$


$\displaystyle = \displaystyle \int_0^1 \int_t^1 \dfrac{\ln^2 z}{z} \, dz \, dt = \displaystyle \int_0^1 -\ln^3 t \, dt = 6 \neq 1$


What was wrong?

$\displaystyle \displaystyle\int \frac{ln(x)}{x}dx$

$\displaystyle \displaystyle = \frac{1}{2}[ ln(x)]^2 +c$

 

[mario99]

$$\displaystyle \int_0^1 \int_t^1 \red{\dfrac{\ln^2 z}{z}} \, dz \, dt$$
This is wrong.

When I was doing the integral, I had teeth-ache and I think that was one of the reasons of forgetting some fragments inside the integral.

$\displaystyle \int_0^1 \int_t^1 \int_z^1 \int_y^1 \dfrac{1}{xyz} \, dx\, dy \, dz \, dt = \displaystyle \int_0^1 \int_t^1 \int_z^1 \dfrac{-\ln y}{yz} \, dy \, dz \, dt$


$\displaystyle = \displaystyle \int_0^1 \int_t^1 \dfrac{\ln^2 z}{2z} \, dz \, dt = \displaystyle \int_0^1 \frac{-\ln^3 t}{6} \, dt = \frac{6}{6} = 1$

I was surprised when I saw another method worked well to solve the integral.

$\displaystyle \int_0^1 \int_0^x \int_0^y \int_0^z \dfrac{1}{xyz} \, \ dt\, dz \, dy \, dx = 1$

This integral is a lot easier to solve. What is the geometry of setting the limits of your integral and this huge integral? Can you show me that?

Why is it wrong to write it like this?

$\displaystyle \int_y^1 \int_z^x \int_t^y \int_0^z \dfrac{1}{xyz} \, \ dt\, dz \, dy \, dx = 1$

 

[mario99]

$\displaystyle \displaystyle\int \frac{ln(x)}{x}dx$

$\displaystyle \displaystyle = \frac{1}{2}[ ln(x)]^2 +c$

I know this. I did the integration mentally and when I transmitted the solution from my brain to Latex code, I forgot to include the numbers in the numerator. Also a strong teeth-ache has distracted me.

 

[BigBeachBanana]

Why is it wrong to write it like this?

$\displaystyle \int_y^1 \int_z^x \int_t^y \int_0^z \dfrac{1}{xyz} \, \ dt\, dz \, dy \, dx = 1$

See post#2 for explicit statements of the domains. Your limits are too loose.

 

[mario99]

See post#2 for explicit statements of the domains. Your limits are too loose.

I have seen it and I still don't understand how to set the limits of integration when the domain is written in series like:

$0 < x < y < z < 1$

 

[BigBeachBanana]

I have seen it and I still don't understand how to set the limits of integration when the domain is written in series like:

$0 < x < y < z < 1$

Start with 2-dimension.
$0 < x < y < 1$

How would set up the limits for this?

 

[mario99]

Start with 2-dimension.
$0 < x < y < 1$

How would set up the limits for this?

$\displaystyle \int_0^y \int_x^1 f(x,y) \ dy \ dx$

 

[BigBeachBanana]

$\displaystyle \int_0^y \int_x^1 f(x,y) \ dy \ dx$

You know that it's wrong since the outermost integral isn't iterated over constants. Explain why 0 ->y.

 

[Dr.Peterson]

$\displaystyle \int_0^y \int_x^1 f(x,y) \ dy \ dx$

Hint:

​

 

[mario99]

You know that it's wrong since the outermost integral isn't iterated over constants. Explain why 0 ->y.

because the domain of $x \ \text{is}: \ 0 < x < y$

Hint:

View attachment 36649​

Thank you Dr.Peterson for passing by. So the idea is that we are taking the area inside a triangle? Why?

 

[Dr.Peterson]

Thank you Dr.Peterson for passing by. So the idea is that we are taking the area inside a triangle? Why?

The hint is for you to think about it. And it's not about area. It's about how you determine limits of integration.

So, what ideas does this give you?

By the way, you could also have dx on the inside, rather than dy. The picture is only one way to do it, intended to match what you did and correct it.

 

[mario99]

The hint is for you to think about it. And it's not about area. It's about how you determine limits of integration.

So, what ideas does this give you?

By the way, you could also have dx on the inside, rather than dy. The picture is only one way to do it, intended to match what you did and correct it.

You are brilliant Dr.Peterson. I tried this hint in problems which have similar domain and it worked. At least, now I have a way to visualize setting the limits of integration.

 

[Dr.Peterson]

You are brilliant Dr.Peterson. I tried this hint in problems which have similar domain and it worked. At least, now I have a way to visualize setting the limits of integration.

Interesting. Did you not know this?

I would expect any chapter introducing double integrals to have a picture similar to this. For example, this textbook section does the same (just lacking the arrowheads, which represent what a teacher might indicate with hand motions while looking at the picture).

(Of course, your original problem in this thread would require a four-dimensional picture, but the idea is the same, and just takes a little stretching of the mind to "visualize".)

 

[mario99]

Of course I know this when I work with two variables, say, $x \ \text{and} \ y$. Using the same concept for more than two variables is something outside the box. It is really amazing to be able to use it there.