Calculate joint conditional distribution of $f(x,y,z) = \frac{xyz}{108}$

Question

Calculate joint conditional distribution of $f(x,y,z) = \frac{xyz}{108}$ for $x = 1, 2, 3; y = 1, 2, 3; z = 1, 2$, of the distribution of $Y$ and $Z$ given $X = 3$.

I first calculated the marginal disitribution for $X$ as $P_X(x) = \frac{x(1)(1)}{108}+\frac{x(1)(2)}{108}+\frac{x(2)(1)}{108}+\frac{x(2)(2)}{108}+\frac{x(3)(1)}{108}+\frac{x(3)(2)}{108}+\frac{x(3)(3)}{108} = \frac{x}{4}$

Then the intersection when $X = 3$ so $P(Z \cap Y \cap X = 3) = \frac{3zy}{108}$

Placing this into the conditional disitribution: $$P(Z \cap Y|X = 3)=\frac{\left(\frac{3zy}{108} \right)}{\left(\frac{3}{4} \right)} = \frac{zy}{27}$$

However, the solutions show: $$\frac{\left(\frac{3zy}{108} \right)}{\left(\frac{1}{2} \right)} = \frac{zy}{18}$$

Is there a mistake with my calculation?

Answer 1

Note that if $p_X(x)=\frac{x}{4}$, you have $1=p_X(1)+p_X(2)+p_X(3)=\frac{1}{4}+\frac{2}{4}+\frac{3}{4}=\frac{3}{2}$, but $1\neq \frac{3}{2}$.

The mistake is in the last term of the sum, $\frac{x(3)(3)}{108}$ is wrong because $3\notin range(Z)$.

Answer 2

It is good practice to use consistent notation. You were given the joint probability mass function $f(x,y,z)$, so you should use similar notation for the marginals and conditionals.

So you would have: $f_{\small X}(x)=\sum_{y,z} f(x,y,z)$ and $f_{\small Y,Z\mid X}(y,z\mid 3)=\left.{f(3,y,z)}\middle/{f_{\small X}(x)}\right.$

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$\qquad\begin{align}f_{\small X}(x)&=\sum_{y,z}f(x,y,z)\\[1ex] &= \dfrac{x}{108}((1+2+3)(1+2))&&\Big[~x\in\{1,2,3\}~\Big]\\[1ex]&=\dfrac{18x}{108}\\[1ex]&=\dfrac x 6\\[2ex]f_{\small Y,Z\mid X}(y,z\mid 3) &=\dfrac{f(3,y,z)}{f_{\small X}(3)}&&\Big[~y\in\{1,2,3\}, z\in\{1,2\}~\Big]\\[1ex]&=\dfrac{3yz/108}{3/6}\\[1ex]&=\dfrac{yz}{18}\end{align}$