# Probability from other probability [closed]

Let $m$, $n$ be very big natural numbers, s.t. $m\leq n$. Let $L\geq 1$ and $Ln\ge 1$. Let also $t>0$, $C>0$ and for some random variable $x$ the following is true: $P(x^2\geq Ln )\geq \frac{C}{(Ln)^t}$.

Show: if $t\geq 4$, then $\frac{n}{2}P(x^2\geq Ln )\leq \frac{1}{m}$. If $t\leq 4$, then $\frac{n}{2}P(x^2\geq Ln )\geq \frac{1}{m}$.

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Is there a typo? $t$ seems to be unrelated. – Matt N. Dec 2 '11 at 8:41
@Matt it appears in the denominator of $P(x^2\geq Ln) \geq C/(Ln)^t$ – Chris Taylor Dec 2 '11 at 9:06
@ChrisTaylor: I need glasses! Thanks Chris. – Matt N. Dec 2 '11 at 9:12

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The level of precision and clarity of this question are well below what is needed for it to be addressed succesfully. At present the hypothesis seems to be that $P(x^2\ge k)\ge C/k^t$ and the desired conclusion when $t\ge4$ seems to be $P(x^2\ge n)\le2/n^2$. There is no way the former (bounding the tail from below) may imply the latter (bounding the tail from above). – Did Dec 2 '11 at 12:40

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David, at least three of us have commented that your question is seriously in need of clarification. Your inattention strikes me as sufficient reason to close the question. – Gerry Myerson Dec 3 '11 at 8:57
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## closed as not a real question by Did, t.b., Gerry Myerson, J. M., SashaDec 3 '11 at 14:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
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