2

I am currently taking a probability course based on the book A first course in probability by Sheldon Ross. I have been trying to solve the following problem:

$f_{X,Y,Z}(x,y,z) = c$ where $x = 1, 2, ..., y \hspace{5mm} y = 1, 2, ..., z \hspace{5mm} z = 1,..., 10 $

Find $\hspace{1mm} f_X(x)$ and $\hspace{1mm} f_Y(y)$

Now, I managed to get the value of $c$ by solving $\sum_{z = 1}^{10} \sum_{y = 1}^z \sum_{x = 1}^y c = 1$. My result was $c = \frac{1}{220}$. However, I cannot get the marginal distributions right. I have done the following: $$f_X(x) = \sum_{z = 1}^{10} \sum_{y = 1}^{z} \frac{1}{220} = \sum_{z = 1}^{10} \frac{z}{220} = \frac{1}{220} * \left( \frac{1}{2}10 *(10+1) \right) = \frac{110}{440} = \frac{1}{4} \hspace{5mm} x = 1,2, ..., 10$$ However, if I sum $f_X$ over all the values of $x$, it obviously does not add up to one. I am having the same issue with the marginal distribution for $y$. I am thinking of doing each individual case ($\sum_{y, z} P[X = 1, Y = y, Z = z]$) to try and grasp the general form, but it seems like it would be a ton of work and there must be an easier way to do it. Am I writing the range of $x$ and $y$ incorrectly or am I adding things wrong or is my joint distribution incorrect altogether?
Thank you for your help!

Community
  • 1
Andres
  • 53

1 Answers1

0

The limits of summation are the problem. They need to be set to cover exactly the possible values each RV can take given a particular $x$ or $y$.

\begin{eqnarray*} f_X(x) &=& \sum_{z=x}^{10}\sum_{y=x}^{z} \dfrac{1}{220} \\ &=& \dfrac{1}{220} \sum_{z=x}^{10}{z-x+1} \\ &=& \dfrac{1}{220} \left((11-x)(1-x) + \dfrac{10\times 11}{2} - \dfrac{x(x-1)}{2} \right) \\ &=& \dfrac{(11-x)(12-x)}{440} \qquad\qquad\qquad\qquad\text{after simplification.} \\ && \\ && \\ f_Y(y) &=& \sum_{z=y}^{10}\sum_{x=1}^{y} \dfrac{1}{220} \\ &=& \dfrac{1}{220} \sum_{z=y}^{10}{y} \\ &=& \dfrac{y(11-y)}{220}. \end{eqnarray*}

Mick A
  • 10,208