# Finding the discrete distribution

I want to find the discrete distribution of X, where expected value of x, E(X)=3 and the Variance of X is 15. X=1,2,3,4,5,and 6. What is the easiest way to find the distribution of X. I really appreciate your response.

• Hint: there is a very simple such distribution with an expectation of $3$ and variance of only $6$: it assigns probability $3/5$ to $1$ and $2/5$ to $6.$ What does that tell you about the number of solutions to your problem? Also see stats.stackexchange.com/questions/45588 and related questions.
– whuber
May 2, 2018 at 22:03
• But it is not a unique solution. May 3, 2018 at 2:29
• @Michael Yes it is.
– whuber
May 3, 2018 at 17:15

$E[X] = 3$ and $Var[X] = 15$ is the same as writing the following equations:

\begin{align} 3 &= E[X] \\ &= \sum_{k=1}^{6} kp_k \\ &= 1p_1 + 2p_2 + 3p_3 + 4p_4 + 5p_5 + 6p_6 \\ \\ \\ 15 &= Var[X] \\ &= E[X^2] - E[X]^2 \\ &= \sum_{k=1}^{6} k^2p_k - (3)^2\\ &= 1^2p_1 + 2^2p_2 + 3^2p_3 + 4^2p_4 + 5^2p_5 + 6^2p_6 -9 \\ \\ \\ 1 &= p_1 + p_2 + p_3 + p_4 + p_5 + p_6 \\ \\ \end{align}

You have 3 equations but 6 unknowns, which means you have an underdetermined linear system. You need 6 equations to solve for 6 unknowns.

note: there is also an inequality constraint that must be satisfied: $$0 <= p_i <= 1 \quad\quad\quad i=1..6$$.

• There is one more equation. $p_1$ + $p_2$ +$p_3$ + $p_4$ + $p_5$ + $p_6$ =1. But you do still need 3 more linear equations in the six unknowns. May 3, 2018 at 2:26
• Thank you very much for the help. Does this mean we need to guess some values for p_i and find the remaining values? Is this the only way to find this? Someone told me about a method called "Generating Functions". Do you know about this approach as well? I really appreciate your time. May 3, 2018 at 9:46
• (-1) This answer is misleading because it omits the crucial inequalities that must be satisfied by the $p_i.$ Consequently, the concept of "underdetermined" is only partly helpful and does not justify the conclusion that you "need 6 equations."
– whuber
May 3, 2018 at 17:15
• @Pradeep - Since the system is underdetermined, there are either no solutions or infinite solutions. In this case it's easy to prove there's at least 1 solution so there are infinite solutions. May 3, 2018 at 17:22
• Since you claim "there's at least 1 solution," would you please exhibit one?
– whuber
May 3, 2018 at 17:22