# 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
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
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