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m In the opening paragraph, I have inserted the SI units in brackets in one of the sentences (SI units J/mol)
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[[File:Markov Inequality.svg|thumb|300px|right|Markov's inequality gives an upper bound for the measure of the set (indicated in red) where <math>f(x)</math> exceeds a given level <math>\varepsilon</math>. The bound combines the level <math>\varepsilon</math> with the average value of <math>f</math>.]]
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In [[probability theory]], '''Markov's inequality''' gives an [[upper bound]] for the [[probability]] that a [[non-negative]] [[function (mathematics)|function]] of a [[random variable]] is greater than or equal to some positive [[Constant (mathematics)|constant]]. It is named after the Russian mathematician [[Andrey Markov]], although it appeared earlier in the work of [[Pafnuty Chebyshev]] (Markov's teacher), and many sources, especially in [[Mathematical analysis|analysis]], refer to it as [[Chebyshev's inequality]] (sometimes, calling it first Chebyshev inequality, while referring to the [[Chebyshev's inequality]] as second Chebyshev's inequality) or [[Irénée-Jules Bienaymé|Bienaymé]]'s inequality.
 
Markov's inequality (and other similar inequalities) relate probabilities to [[expected value|expectation]]s, and provide (frequently loose but still useful) bounds for the [[cumulative distribution function]] of a random variable.
 
An example of an application of Markov's inequality is the fact that (assuming incomes are non-negative) no more than 1/5 of the population can have more than 5 times the average income.
 
==Statement==
If ''X'' is any nonnegative random variable and ''a''&nbsp;>&nbsp;0, then
 
: <math>\mathbb{P}(X \geq a) \leq \frac{\mathbb{E}(X)}{a}.</math>
 
In the language of [[measure theory]], Markov's inequality states that if (''X'',&nbsp;Σ,&nbsp;''μ'') is a [[Measure (mathematics)|measure space]], ''ƒ'' is a [[measurable function|measurable]] [[extended real number line|extended real]]-valued function, and <math>\varepsilon>0</math>, then
 
:<math> \mu(\{x\in X:|f(x)|\geq \varepsilon \}) \leq {1\over \varepsilon}\int_X |f|\,d\mu.</math>
 
(This measure theoretic definition may sometimes be referred to as [[Chebyshev's inequality]]
.<ref>E.M. Stein, R. Shakarchi, "Real Analysis, Measure Theory, Integration, & Hilbert Spaces", vol. 3, 1st ed., 2005, p.91</ref>)
 
==Corollary: Chebyshev's inequality==
 
[[Chebyshev's inequality]] uses the variance to bound the probability that a random variable deviates far from the mean. Specifically:
 
:<math>\mathbb{P}(|X-\mathbb{E}(X)| \geq a) \leq \frac{\textrm{Var}(X)}{a^2},</math>
 
for any ''a>0''. Here Var(X) is the variance of X, defined as:
 
:<math> \operatorname{Var}(X) = \mathbb{E}[(X - \mathbb{E}(X) )^2]. </math>
 
Chebyshev's inequality follows from Markov's inequality by considering the random variable
 
: <math> (X - \mathbb{E}(X))^2 </math>
 
for which Markov's inequality reads
 
: <math> \mathbb{P}( (X - \mathbb{E}(X))^2 \ge a^2) \le \frac{\operatorname{Var}(X)}{a^2}, </math>
 
==Proofs==
We separate the case in which the measure space is a probability space from the more general case because the probability case is more accessible for the general reader.
 
===Proof In the language of probability theory===
 
For any event ''E'', let ''I''<sub>''E''</sub> be the indicator random variable of ''E'', that is, ''I''<sub>''E''</sub> =&nbsp;1 if ''E'' occurs and ''I''<sub>''E''</sub> =&nbsp;0 otherwise.
 
Using this notation, we have ''I''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub>&nbsp;=&nbsp;1 if the event ''X''&nbsp;≥&nbsp;''a'' occurs, and ''I''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub>&nbsp;=&nbsp;0 if ''X''&nbsp;<&nbsp;''a''. Then, given ''a''&nbsp;>&nbsp;0,
 
:<math>aI_{(X \geq a)} \leq X\,</math>
 
which is clear if we consider the two possible values of ''I''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub>. If ''X'' < ''a'', then ''I''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub> = 0, and so ''aI''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub> =&nbsp;0 ≤&nbsp;''X''. Otherwise, we have ''X''&nbsp;≥&nbsp;''a'', for which ''I''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub> = 1 and so ''aI''<sub>(''X''&nbsp;≥&nbsp;''a'')</sub> =&nbsp;a ≤&nbsp;''X''.
 
Since <math>\mathbb{E}</math> is a linear operator, taking expectation of both sides of an inequality cannot reverse it. Therefore
 
:<math>\mathbb{E}(aI_{(X \geq a)}) \leq \mathbb{E}(X).\,</math>
 
Now, using linearity of expectations, the left side of this inequality is the same as
 
:<math>a\mathbb{E}(I_{(X \geq a)}) = a(1\cdot\mathbb{P}(X \geq a) + 0\cdot\mathbb{P}(X < a)) = a\mathbb{P}(X \geq a).\,</math>
 
Thus we have
 
:<math>a\mathbb{P}(X \geq a) \leq \mathbb{E}(X)\,</math>
 
and since ''a'' > 0, we can divide both sides by ''a''.
 
===In the language of measure theory===
 
We may assume that the function <math>f</math> is non-negative, since only its absolute value enters in the equation. Now, consider the real-valued function ''s'' on ''X'' given by
 
:<math>
s(x) =
\begin{cases}
  \varepsilon, & \text{if } f(x) \geq \varepsilon  \\
  0, & \text{if } f(x) < \varepsilon
\end{cases}
</math>
 
Then <math>0\leq s(x)\leq f(x)</math>. By the definition of the [[Lebesgue integral]]
 
:<math>
\int_X f(x) \, d\mu \geq \int_X s(x) \, d \mu = \varepsilon \mu( \{ x\in X : \, f(x) \geq \varepsilon \} )
</math>
 
and since <math>\varepsilon >0 </math>, both sides can be divided by <math>\varepsilon</math>, obtaining
 
:<math>\mu(\{x\in X : \, f(x) \geq \varepsilon \}) \leq {1\over \varepsilon }\int_X f \,d\mu.</math>
 
[[Q.E.D.]]
 
==Matrix-valued Markov==
Let <math> M \succeq 0 </math> be a self adjoint matrix-valued random variable and <math> a>0 </math>. Then
:<math>
\mathbb{P}(M \npreceq a \cdot I) \leq \frac{\mathrm{tr}\left( E(M) \right)}{a}.
</math>
 
==Examples==
* Markov's inequality is used to prove [[Chebyshev's inequality]].
* Markov's inequality can be used to show that, for a nonnegative random variable, the mean <math>\mu</math> and a median <math>m</math> are such that <math>m \le 2 \mu</math>.
 
==See also==
* [[Doob martingale#McDiarmid.27s_inequality|McDiarmid's inequality]]
* [[Bernstein inequalities (probability theory)]]
*[[Chernoff_bound|Chernoff Bound]]
 
== References ==
{{reflist}}
 
==External links==
* [http://mws.cs.ru.nl/mwiki/random_1.html#T36 The formal proof of Markov's inequality] in the [[Mizar system]].
 
{{Refimprove|date=September 2010}}
 
{{DEFAULTSORT:Markov's Inequality}}
[[Category:Probabilistic inequalities]]
[[Category:Articles containing proofs]]

Revision as of 03:30, 27 February 2014

I cannot state which my case is going to be like everyone else's. Following all, no 2 cases are precisely the same. Hemorrhoids could vary from individual to individual and case to case. What I can say, although, is usually which I identified an internal hemorrhoid treatment that worked for me plus that I could claim it is very probably to function for we, also. Let's not get too formal with last names plus such. Right after all, piles aren't a disorder that anybody is really satisfied to have or even discuss. It needs to be debated, though. I desire fellow hemorrhoid sufferers including me to understand how to escape it the means I did thus.

Now that a hemorrhoids aren't getting worse, it's time to use internal hemorrhoid treatment which will relieve itching and pain. Soaking inside a a sitz shower for regarding fifteen minutes at a time is 1 technique to discover relief. We can do this numerous times a day. You are able to moreover heat a wet towel and apply it to the hemorrhoid area. Ice for ten minutes, followed by a warm compress for an equal or somewhat longer length of time is another method to relieve a pain plus itching.

Another treatment for hemorrhoids will be to use witch hazel. It's been shown to succeed at helping decrease swelling and helps with healing hemorrhoids which are bleeding. If your piles are very swollen, this can undoubtedly aid you today.

Hemorrhoid Surgery - This is only suggested if your hemorrhoids are thus serious which no over the counter or house remedy is providing you any sort of relief.

Most of the time you will prefer to stay at house in purchase to do anything with all the pain which you are experiencing. Nowadays hemorrhoids are not really a big deal due to the accessibility of those hemorrhoid treatments. The best treatment for this is to learn the cause of hemorrhoid plus understand the procedures that will help we receive out of which painful condition.

Step 5 - Consume Sufficient Fiber. Your mother can possibly have told we this when a kid and now you'll recognize it's correct. Not consuming an adequate amount of fiber will truth be told cause constipation plus harder stools. A significant fiber diet can help to avert flare ups and keep a stools soft. Fresh fruits, veggies, beans and peanut butter are all splendid sources of fiber to aid keep we standard.

As you are able to see there are different types of hemorrhoid treatment which you can use at home. Some of them can provide a remarkable deal of relief from this uncomfortable condition, thus it does pay to test them out. Do this and you'll soon notice you are no longer having to cope with hemorrhoids on a daily basis.