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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.

Pradeep
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    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 https://stats.stackexchange.com/questions/45588 and related questions. – whuber May 02 '18 at 22:03
  • But it is not a unique solution. – Michael R. Chernick May 03 '18 at 02:29
  • @Michael Yes it is. – whuber May 03 '18 at 17:15

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$E[X] = 3$ 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: $$0 <= p_i <= 1 \quad\quad\quad i=1..6$$.

anthonybell
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    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. – Michael R. Chernick May 03 '18 at 02:26
  • 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. – Pradeep May 03 '18 at 09:46
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    (-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 03 '18 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. – anthonybell May 03 '18 at 17:22
  • Since you claim "there's at least 1 solution," would you please exhibit one? – whuber May 03 '18 at 17:22
  • Actually with the probability constraints it is not easy to prove. I ran it through a linear solver but the solver couldn't find a feasible starting point, so I guessing there's a fairly good chance there is actually no solution. – anthonybell May 03 '18 at 18:02
  • There's yet one more inequality constraint that must be satisfied. It is indicated in comments to the question: the variance cannot exceed $5^2(3/5 \times 2/5).$ What solutions can you find that satisfy this? – whuber May 04 '18 at 19:00
  • The comment about "variance of a random variable" is an upper bound that is derived from the mean and variance equations so it does not need to be added as an explicit constraint but... for this specific problem that upper bound proves there is no solution! That's a neat proof, thanks for sharing! – anthonybell May 04 '18 at 21:01