There's two problems in my book that utilize a somewhat similar technique for describing probabilities of random variables taking on a value.
Example 1
Random variable $X$ never takes on a value greater than $10$. During a proof using Chebyshev's inequality, this statement confused me.
$Pr[X \geq 6] \leq Pr[|X-2| \geq 4]$ because $X \geq 6$ is a subset of the event $|X-2| \geq 4$
I'm unconvinced by $Pr[X \geq 6] \leq Pr[|X-2| \geq 4]$. Since they used the word "subset" I did the following:
Out of 10 values, these are the only values $X$ can take on such that $X-2 \geq 4$: 10, 9, 8, 7, 6.
Out of 10 values, these are the only values $X$ can take on such that $X \geq 6$: 10, 9, 8, 7, 6
They both seem to have the same values, so by subset did they mean strict subset? If so then when would the "less than" part happen? They would always be equal. I'm probably thinking about this the wrong way.
Example 2
Random variable $X$ takes on a value between $0$ and $70$. It represents a score in a class. You get an A if you score $60$ or above. $Var(X) = 25$ and $E(X) = 35$. Use Chebyshev's inequality o conclude that you have less than $5$% chance of getting an A.
$Pr[X \geq 60] \leq Pr[|X-35| \geq 25] \leq \frac{Var(X)}{25^2} = \frac{1}{25}$
I understand that $\mu = 35$ but why did they set $|X-35| \geq 25$ ? This also seems to use a similar technique to the first example.
For example 1, you forget that $|X-2|\ge 4$ also contains the possible values $-2,-3,\ldots.$ Assuming the only possible values are $1,2,3,4,5,\ldots 9,10$ then you aren't really thinking about it the wrong way... it would be true in that case that $P(X\ge 6) = P(|X-2|\ge 4).$ Note that in this case it is also true that $P(X\ge 6) \le P(|X-2|\ge 4).$
But it's not always going to be the case that the only possible values are between one and ten. A lot of times any real number is possible. In those cases it will still be true that the event $X\ge 6$ is a subset of the event $|X-2|\ge 4$ and this will have the consequence that $P(X\ge 6) \le P(|X-2|\ge 4).$ And it won't necessarily be the case that they are equal since those negative possibilities could have positive probability.
For the second example they moved from $P(X\ge 60)$ to the form $P(|X-35|\ge 25)$ because they wanted to use Chebyshev's theorem, which applies to the probability the deviation from the mean exceeds a certain amount. They have ascertained that the event $X \ge 60$ has the same probability as $|X-35| \ge 25.$ And yes, it's the same as the last example. $|X-35|\ge 25$ implies that $X\ge 60,$ so the event $X\ge 60$ is a subset of $|X-35|\ge 25.$ This means that whenever the first holds, the second will as well, and thus the second has a greater than or equal probability.