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I have the random variable X that returns the result the thrown of one dice such that: $P(1) = P(3) = P(4) = p_1, \\ P(5) = P(6) = p_2, \\ P(2) = \frac{1}{2}P(5)$
Determine the values of $p_1$ and $p_2$:

I have considered the definition of discrete random variable:
$$\sum_{\forall x} P(X=x) = 1 $$ therefore I have done:

$p_1 + \frac{1}{2}p_2 + p_1 + p_1 + p_2 + p_2 = 1 \\ 3p_1 + \frac{5}{2}p_2 = 1$

by attempts I have found that possible values can be:
$p_1 = \frac{1}{3} \land p_2=0$,
$p_1 = 0 \land p_2 = \frac{2}{5}$

Do exist other possible values?
Is there any algorithm to calculate that values without attempts?
Thanks!

JB-Franco
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1 Answers1

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The constraints that need to be satisfied are $$3p_1+ \frac{5}{2}p_2 =1 $$

and $$p_1 \geq 0$$ $$p_2 \geq 0$$

The constraint set is convex.

Hence any convex combination of two solution is a solution.

$$(p_1, p_2) = \lambda (\frac13, 0) + (1-\lambda) (0, \frac25)$$

where $\lambda \in [0,1]$.

Remark: if you sketch out the feasible set, you will find that the feasible set is just a line segment where you have found the end points.

Siong Thye Goh
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