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I need to compute 2 discrete distributions with equal mean and variance. One should be a distribution taking values x and y and the other one taking values z, 0, and y. The probabilities should take the following form:

  • Distribution 1: takes values x and y. prob(x)=0.3; prob(y)=0.7
  • Distribution 2: takes values y, z and v with Prob(y)=0.7; prob(v)=q; prob(z)=r. Y=6000; V=0.

The idea is to choose z and x so that both distributions have equal mean and variance.

I'm not an expert so I don't know if there are too many unknowns or if this is possible at all. What parameters could I change to make a solution possible? I would be very grateful for hints how to solve this.

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    $\begingroup$ Is this for a class? $\endgroup$
    – Glen_b
    Commented Feb 6, 2015 at 9:22
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    $\begingroup$ could you write down both equations that all the unknowns have to statisfy for getting equal mean and equal variance? $\endgroup$
    – Xi'an
    Commented Feb 6, 2015 at 9:52
  • $\begingroup$ no it's not for a class, it is for an experiment I want to design. unfortunately I am not familiar with mathematical software which should solve this easily. The equations are the following: $\endgroup$
    – hanna
    Commented Feb 6, 2015 at 11:37
  • $\begingroup$ Mean=μ=3/10 X=p_zZ+p_vV p_z+p_v=3/10 Variance= σ^2=1/2 [(X-μ)^2+(Y-μ)²]=1/3 [(X-μ)^2+(Z-μ)^2+(V-μ)²] V=0;Y=6000 $\endgroup$
    – hanna
    Commented Feb 6, 2015 at 11:47
  • $\begingroup$ You need to revise your problem statement again. Does random variable 2 take on values $x,$ $z,$ and $v?$ Or is it $y,$ $z,$ and $v?$ $\endgroup$
    – soakley
    Commented Feb 7, 2015 at 17:33

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Let the first (two point) random variable be $A$ and the second be $B.$ Let $q$ be the probability that $B$ takes on the value $0.$ Instead of $r$ as in your problem statement, we equivalently let the probability that $B$ is $z$ be $0.3-q.$ Equating the means, we have $$0.3x+0.7(6000) = (0.3-q)z+0.7(6000),$$

so $$x = {z \left( 1-{q \over 0.3} \right) }$$

If the variances are identical, then the second moments about the origin are the same also. This gives us

$$E[A^2]= 0.3x^2 + 0.7 (6000)^2 = (0.3-q)z^2 + 0.7(6000)^2$$

Plugging in our expression for $x,$ this leads to $$0.3 \left[z \left( 1 - {q \over 0.3} \right) \right]^2 = 0.3z^2 -qz^2$$ Solving this for $q$ gives the two solutions $q=0$ or $q=0.3,$ both of which imply that $A$ and $B$ have the same two-point distribution.

So your problem as posed does not have a solution in terms of your definition of an acceptable answer.

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  • $\begingroup$ Thank you for your comment. But Ithink this misses out that the second distribution should take 3 different values, one takes value 0, one takes value 6000 and one is unknown. This woud change the variance of the second distribution. $\endgroup$
    – hanna
    Commented Feb 7, 2015 at 9:03
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    $\begingroup$ What it shows is that there was no solution that had positive probabilities for all three outcomes of the random variable $B.$ But now you have changed the problem definition again. $\endgroup$
    – soakley
    Commented Feb 7, 2015 at 13:52
  • $\begingroup$ Note that this does not say that you cannot have a 2-point distribution and a 3-point distribution with equal mean and variance. Just not under your constraints. So, for example, consider that $A$ takes on $-6$ with probability ${1 \over 2}$ and $+6$ with probability ${1 \over 2}.$ Then let $B$ take on $-12$ with probability ${1 \over 8},$ zero with probability ${3 \over 4},$ and $+12$ with probability ${1 \over 8}.$ These have identical means and variances. $\endgroup$
    – soakley
    Commented Feb 10, 2015 at 23:01
  • $\begingroup$ Daer Soakley, is there any sort of formula that tells me under which constraints it is possible? The main constraint I want to implement is that the probability for the highest outcome should be the same in both cases. $\endgroup$
    – hanna
    Commented Feb 11, 2015 at 13:32
  • $\begingroup$ No formula that I know about. I don't think it is possible to do what you want even under only the main constraint. Think of it this way. Since you are leaving the highest point the same for both cases, you need to take the lower point of the 2-point distribution and divide it into 2 pieces (to make the 3-point distribution). You can keep the mean the same by dividing that point symmetrically. But the variance will then have to increase since you are including a point further to the left. That's far from rigorous, but experiment around a bit to convince yourself. $\endgroup$
    – soakley
    Commented Feb 11, 2015 at 23:33

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