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Is this equality generally true? | Probability

  1. Oct 27, 2010 #1
    Suppose Xn is a random variable. Let b and c be a constant.

    Is the following generally true?

    [tex]P(|X_{n}-b| \geq \epsilon) = P(|X_{n}-b|^{2} \geq \epsilon^{2}) [/tex]

    This says that the probability that Xn minus b is greater than or equal to epsilon is equal to the probability that Xn minus b squared is greater than epsilon squared.

    My prof keeps saying that they are the same event, therefore, they have the same probability. But I still don't understand. Any insight?

    Thanks,
    M
     
  2. jcsd
  3. Oct 27, 2010 #2

    jbunniii

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    Yes, it's generally true. It is because for any positive numbers [itex]a[/itex] and [itex]b[/itex],

    [tex]a \geq b[/tex]

    if and only if

    [tex]a^2 \geq b^2[/tex]
     
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