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Roll a pair of four-sided dice, one red and one black. Let $X$ equal the outcome on the red die and let $Y$ equal the sum of the two dice. Define the joint pmf on the space.

So far I have $X = 1,2,3,4$ and $Y = 2,3,4,5,6,7,8$. Each outcome on each die has a $\frac{1}{4}$ probability of being rolled and thus each outcome of the combined rolls is $\frac{1}{16}$. There are two ways to make each value in $Y$. For example to make $3$ we could have $2$ on the red die and $1$ on the black or $1$ on the red die and $2$ on the black die. $\frac{1}{16} + \frac{1}{16} = \frac{2}{16}$. Therefore the pmf should = $\frac{2}{16}$. This is not the answer, however :). Ideas?

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    $\begingroup$ Usually "pmf" means "probability mass function." A single number like $2/16$ is not such a function, so what exactly do you mean by "pmf"? Questions very similar to this one have many answers on this site, so linking to that search would give you a lot of relevant information. $\endgroup$
    – whuber ♦
    Commented Dec 12, 2014 at 22:27
  • $\begingroup$ A constant is a perfectly valid value for a pmf, just as a constant is a valid value for any function that does not depend on the variable(s) in question. In this case we observe that the probability of the distribution is constant over all values of $X$ in the set. $\endgroup$
    – Chris
    Commented Dec 12, 2014 at 22:46
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    $\begingroup$ I'm afraid it's not possible even to determine what this set is from your question. What PMF are you actually trying to compute? For $X$, $Y$, $X+Y$, or something else? It's not so obvious because if $Y$ really does represent the sum of two dice, there clearly are not "two ways to make each value"; for instance, the only way for the sum to equal $2$ is for both the dice to show $1$. Thus your descriptions of $Y$ are contradictory. $\endgroup$
    – whuber ♦
    Commented Dec 12, 2014 at 23:18
  • $\begingroup$ The pmf under question is that of the bivariate distribution that depends on both $X$ and $Y$. Each variable has a set (as described in the question above) of possible values based on their descriptions of rolling two four-sided die ($X$ as the red die's outcome, $Y$ as the sum of the black and red die). The pmf of a bivariate (or any) distribution describes the probability (often but not always dependant on variables) of each outcome. I've actually already (correctly) answered this question below, but for what it's worth if you roll two of any die there are always two ways to make each sum. $\endgroup$
    – Chris
    Commented Dec 12, 2014 at 23:25
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    $\begingroup$ It's impossible to tell whether an answer is correct when the question cannot be understood! $\endgroup$
    – whuber ♦
    Commented Dec 12, 2014 at 23:29

1 Answer 1

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Writing this up led me to the solution.

When we plug in values into the pmf $f(x,y)$ such as $f(1,2)$ we must recall that we have already decided what the value of $X$ is. Although there are two ways to sum two four-sided to get the value $2$, we have already set $X=1$ and therefore there is only one value ($1$) that the remaining die may have to make $2$. Thus there is only $1$ way to create $2$ given the parameters of the pmf and thus the probability for this and all rolls is static.

$f(x,y) = \frac{1}{16}$

Table

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  • $\begingroup$ Thank you for quoting the original question (+1). It asks for a joint probability mass function (pmf). By definition this will assign a probability to each ordered pair $(x,y)$ where $x$ is one of $1,2,3,4$ and $Y$ is one of $2,3,4,5,6,7,8$. Altogether these constitute $4\times 7=28$ values you need to specify. Clearly they cannot all equal $1/16$, because they must sum to $1$. $\endgroup$
    – whuber ♦
    Commented Dec 13, 2014 at 0:29
  • $\begingroup$ Continuing whuber's comment, what is the probability of $f(1,8)$? or of $f(2,2)$? $\endgroup$
    – Alecos Papadopoulos
    Commented Dec 13, 2014 at 3:31
  • $\begingroup$ Whuber: You cannot simply multiply the possible values of $X$ and $Y$ as some values cannot be associated with each other. For example if $X=2$ (the red die shows a value of 2) it is impossible for $Y$ (the sum of the two die) to be 2 as the black die will be 1,2,3, or 4 and the sum of the two die must be greater than 2. See the table above. Note there are 16 possible values and thus the sum is 1. $\endgroup$
    – Chris
    Commented Dec 13, 2014 at 14:10
  • $\begingroup$ Alecos: $f(1,8)$ asks: If the red die shows a value of 1, what is the probability that the sum of the red and black die is 8? It is not possible for the sum of two four-sided die, one showing a 1, to equal 8. Thus this is undefined. See my comment to whuber for $f(2,2)$. $\endgroup$
    – Chris
    Commented Dec 13, 2014 at 14:15
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    $\begingroup$ It is not undefined Chris, it is zero. The support of the two variables is determined as the Cartesian product of their individual supports. Of course, since the probability for some pairs is zero, the shortcut is to "ignore" these values and present only the rest. It won't affect any subsequent calculations, but the correct derivation is to show all possible values and assign zero probability where it is due -Hogg et al. textbook notwithstanding. $\endgroup$
    – Alecos Papadopoulos
    Commented Dec 13, 2014 at 17:34

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