This machine is predictable

Probability Level 3

I have a machine that follows a probability distribution and it only spits out a value of 1 or -1.

If the mean and variance of this probability distribution are equal, is it more likely for the machine to spit out a value of 1 or a value of -1?

It is more likely to spit out a value of 1 They are both equally likely It is more likely to spit out a value of -1

1 solution

Brian Charlesworth
Jul 15, 2016

Variance, ( $\sigma^{2}$ ), is necessarily a non-negative value, and if both the mean, ( $\mu$ ), and variance were $0$ then the only possible output would be $0$ , which is not the case. Thus the variance, and thus also the mean, are positive, indicating that an output of $1$ must be more likely than an output of $-1$ .

More rigorously, given that $\sigma^{2} = \mu$ , and noting that $E(X^{2}) = 1$ since $X^{2} = 1$ regardless of whether the output is $-1$ or $1$ , the formula $\sigma^{2} = E(X^{2}) - \mu^{2}$ becomes

$\mu = 1 - \mu^{2} \Longrightarrow \mu^{2} + \mu - 1 = 0 \Longrightarrow \mu = \dfrac{-1 \pm \sqrt{5}}{2}$ .

As noted before, since variance is necessarily non-negative, we can conclude that $\mu = \dfrac{\sqrt{5} - 1}{2} = \phi - 1$ .

Challenge Master Note: Wonderful interpretation of this problem! For completeness, you should show that there exists probabilities in interval $[0,1]$ for these two outputs.

Pi Han Goh - 4 years, 11 months ago

Let $P(1) = p$ . Then $P(-1) = 1 - p$ and so

$p*(1) + (1 - p)*(-1) = \mu \Longrightarrow 2p = \mu + 1 \Longrightarrow p = \dfrac{\mu + 1}{2} = \dfrac{\sqrt{5} + 1}{4} \approx 0.809$ ,

and thus $P(-1) = 1 - p = \dfrac{3 - \sqrt{5}}{4} \approx 0.191$ .

Brian Charlesworth - 4 years, 11 months ago

This machine is predictable

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