Finding a distribution

Question

Hey, I'm having trouble understanding the following question:

Consider the random variable X that can take on the values
1, 2, 3, 4, and 5. Assume that the numbers are equally
likely. Hence the expected value of X equals 3.

Now, find a distribution P∗ for X such that for at least two
elements i, j of the set {1, . . . , 5} we have
P∗(X = i) is NOT equal to P*(X = j), but nevertheless the summation of k=1 to k=5 of k*P(X=k) = 3.

Answer 1

Any solution in [0,1]⁵ to the following system is a distribution:

 

p₁ + p₂ + p₃ + p₄ + p₅ = 1

 

p₁ + 2p₂ + 3p₃ + 4p₄ + 5p₅ = 3

 

we can choose any values for p₃ , p₄ , and p₅ and then solve uniquely for p₁ and p₂

 

Specifically:

 

Subtract equation 1 from equation 2 to and solve for p₂ to get:

 

p₂ = 2 - 2p₃ - 3p₄ - 4p₅

 

Subtract equation 2 from equation twice 1 to and solve for p₁ to get:

 

p₁ = p₃ + 2p₄ + 3p₅ - 1

 

We have constraints 1 < p₃ + 2p₄ +3p₅ < 2p₃ + 3p₄ + 4p₅ < 2.

 

We already know that p₃ = p₄ = p₅ = 1/5 work since that is from the uniform distribution. here the two expressions in the constraint are equal to 6/5 and 9/5. To make a non-uniform, we can just tweak a bit. For example, letting p₃ = 1/4 makes the expressions have values 5/4 and 19/10.

 

Then we have p₁ = 5/4 - 1 = 1/4 and p₂ = 2 - 19/10 = 1/10

 

Then this is our new distribution:

 

P(X=1) = 1/4

P(X=2) = 1/10

P(X=3) = 1/4

P(X=4) = 1/5

P(X=5) = 1/5