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Given: For some $X$, $Var(X) = 9$, $E(X) = 2$, $E(X^2) = 13$

Problem: $Pr[X = 2] > 0$

The solution in my book says to construct a r.v. to satisfy the above conditions and confirm or deny the statement from there. For simplicity, assume $X$ can take on two values $Pr[a] = \frac{1}{2}$ and $Pr[b] = \frac{1}{2}$. Finding that $a$ and $b$ are not $2$ is enough to disprove the statement. We can apply the constraints:

$\frac{1}{2}a +\frac{1}{2}b = 2$

$\frac{1}{2}a^2 + \frac{1}{2}b^2 = 13$

The book then says to solve for $a$ and $b$. I've never solved a nonlinear systems of equations before, and I'm not sure if we have to to solve this problem. So I naively did this:

$a = 4 -b$ from the first equation

$b^2 - 4b - 5 = 0$ by substituting $a$

The above quadratic has two solutions, $b = -1$ and $b = 5$ which happen to be the same solutions the book got for $a$ and $b$ respectively.

I'm not sure if this is the right way to solve this problem since I never explicitly solved for $a$ and the two solutions for $b$ may coincidentally be the same.

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    $\begingroup$ Before commenting on your solution, a small step back: what does it have to do with Markov's inequality? (where was in mentioned -- in the text of the exercise? Because, as far as I can tell, it is used exactly nowhere) $\endgroup$ – Clement C. May 30 '17 at 14:22
  • $\begingroup$ @ClementC. Oops thanks for pointing this out. It was under several questions regarding Markov's Inequality so I got mixed up. Changed the title $\endgroup$ – Carpetfizz May 30 '17 at 14:25
  • $\begingroup$ Now, regarding your solution: once you solve for $b$, you have two options, $-1$ and $5$. Using the first equation, this'll give you $(a,b)=(4-(-1),-1)=(5,-1)$ or $(a,b)=(4-5,5)=(-1,5)$. Clearly, both are valid (by your choice of parameterization, the roles of $a$ and $b$ are symmetric: if $(x,y)$ is solution, so is $y,x)$. $\endgroup$ – Clement C. May 30 '17 at 14:26
  • $\begingroup$ @ClementC. Thanks that makes sense $\endgroup$ – Carpetfizz May 30 '17 at 14:27
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You haven't made a mistake.

Once you solve for $b$, you have two options, $-1$ and $5$. Using the first equation now, you'll get the corresponding two possibilities for $a$, since $$ a=4-b $$ This will give you $$(a,b)=(4-(-1),-1)=(5,-1)\tag{1}$$ or $$(a,b)=(4-5,5)=(-1,5)\tag{2}$$ Clearly, both are valid (by your choice of parameterization, the roles of $a$ and $b$ are symmetric: if $(x,y)$ is solution, so is $(y,x)$.


Small comments: the statement of the exercise is redundant, as if you have $\mathbb{E}[X^2]$ and $\mathbb{E}[X]$, you already know the variance as $\mathrm{Var} X = \mathbb{E}[X^2]-\mathbb{E}[X]^2$.

Further, for simplicity I would have personally solved it with a random variable taking two values, $0$ and $\alpha$, (respectively with probability $1-p$ and $p$), and solved for $(\alpha,p)$. This gives a different counterexample — try it out.

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  • $\begingroup$ Thanks again, will accept once it lets me $\endgroup$ – Carpetfizz May 30 '17 at 14:28
  • $\begingroup$ @Carpetfizz I've added two small comments, to clarify a couple points. $\endgroup$ – Clement C. May 30 '17 at 14:30
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    $\begingroup$ Ah yeah, I added those as given, as I had already solved for $E(X^2)$ using the equality you have provided. Will try the new method and comment here if I get stuck! $\endgroup$ – Carpetfizz May 30 '17 at 14:33

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