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It is given that $E[X^n] = \frac{2}{5}(-1)^n + \frac{2^{n+1}}{5}+\frac{1}{5}$, where $n=1,2,3,\ldots.$ I need to find $P(|X-\frac{1}{2}| > 1)$.

What my approach is : I have opened the modulus inequality and equated this probability to $1-P(\frac{-1}{2}<X<\frac{3}{2})$. This can only occur when X=0 and X=1, so that's what I need to find. So, I thought I can find the MGF using the expectation and expanding it, that is, $ M_x(t) = \sum_{n=1}^{\infty} \frac{t^n}{n!}E(X^n).$ And then use it to find the PMF, or corresponding probabilities directly from the summation that I've obtained.

However, I'm stuck here and not able to proceed. Is my approach wrong? If no, then how should I proceed?

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    $\begingroup$ The data is consistent with a 2/5 chance that X=-1, a 2/5 chance that X=2, and 1/5 chance that X=1, in which case the given probability is 4/5. Furthermore, it should be possible to show that this is the only possible distribution by proving $P[|X|]>2]=2/5, P[|X|\ge 1]=1$, etc $\endgroup$
    – Matt F.
    Commented Mar 8 at 2:56
  • $\begingroup$ Okay, but how should I go about and figure that out? $\endgroup$ Commented Mar 8 at 3:01
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    $\begingroup$ To the close voter, I think this is a genuine conceptual query and OP showed their approach on solving this. $\endgroup$ Commented Mar 8 at 13:17

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\begin{align}E[X^n]&=\frac25(-1)^n + \frac{2^{n+1}}{5}+\frac15\\ &=\frac25\cdot (-1)^n + \frac25\cdot 2^{n}+\frac15\cdot (1)^n\end{align}

From this expression, I can see that $P(X=-1)=\frac25$. Can you identify the probability of $P(X=2)$ and also the other value that $X$ takes with probability $\frac15$?

With that you should be able to solve the problem of your interest.

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You can consider

$$M_x(t) = \sum_{n=1}^{\infty} \frac{t^n}{n!}E(X^n) = \sum_{i=1}^{m} e^{tx_i} p_i$$

Or more directly

$$E[X^n] = \sum_{i} p_i x_i^n $$

You have three terms that are powers of $n$ and they can be explained with three point masses

$$E[X^n] = p_1 x_1^n + p_2 x_2^n +p_3 x_3^n$$

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  • $\begingroup$ Got it. I was overcomplicating the problem by using the MGF approach $\endgroup$ Commented Mar 8 at 9:58

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