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Let $Y$ be a random variable such that $E[Y] = \lambda$, $\lambda \in \mathbb{R}$ and $E[Y^2]<\infty$. The problem is to find a lower bound on the probability $$ P \left[|Y| > \frac{|\lambda|}{2} \right]. $$

Any leads would be appreciated.

DifferentialPleiometry
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Bhisham
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  • Do you have knowledge on the $\mathbb{E}[|Y|]$ and $\mathbb{E}[|Y|^{2}]$, because in that case the Paley-Zygmund inequality might be what you want – Fiodor1234 Jul 27 '21 at 09:37

2 Answers2

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Edit : This answer applies to the original question, that was : "for ANY random variable Y, what is the lower bound of (formula)"

The lower bound is 0.

Let's take $(Y_{n})$ a series following a Bernoulli law such as : $P(Y_{n} = 1) =1/n$. Then, ${E(Y_{n})= \lambda = 1/n}$.

$P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2}) = P(\lvert Y_{n} \rvert > \frac{1}{2n}) = P(Y_{n} =1) = \frac{1}{n} < \frac{2}{n}\underset{n\to +\infty}{\longrightarrow} 0 $

Then, $0 < P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2}) < \frac{2}{n}$ , so asymptotically, the lower bound is 0.

Adrien
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    Hello and welcome to Cross Validated! Could you please format your answer using [LaTeX](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)? – mhdadk Jul 27 '21 at 13:25
  • How does a series of different random variables respond to a question about a *single* random variable?? – whuber Jul 27 '21 at 18:47
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    @whuber my answer was posted before the edit of galen. The question was : "let Y ANY random variable". Then, the question is to find a lower bound for all the possible random variables, and my answer with series provides the response : the lower bound is 0 – Adrien Jul 27 '21 at 20:52
  • @mhdadk done it ! – Adrien Jul 27 '21 at 23:58
  • You still misinterpret what "any" means: sure, the RV can be arbitrary: but a sequence of *different* random variables with *different* values of $\lambda$ is not "any" random variable! As such, your analysis sheds no light on the question. If, instead, you were to scale your examples suitably, you would exhibit variables with a *common* value of $\lambda$ and as a *set* (not a series) they would demonstrate the nonexistence of any positive lower bound *for a given $\lambda.$* – whuber Jul 28 '21 at 15:26
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Given $\lambda,$ there must be a universal lower bound $p(\lambda)$ (because $0$ will certainly work.) The question is whether there are any $\lambda$ where this bound exceeds $0.$ Regardless, being a lower bound means that for any random variable $Y$ with $E[Y^2]\lt \infty$ and $E[Y]=\lambda,$ $$\Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).\tag{*}$$

Consider $Y=\lambda X/q$ where $X$ is a Bernoulli$(q)$ variable and $\lambda \ne 0.$ Because $E[X]=q,$ $E[Y] = \lambda (q)/q = \lambda.$ $Y$ also has finite second moment because it is bounded. Moreover, since $q \lt 2,$ the event "$|Y|\gt|\lambda|/2$" equals the event "$Y=\lambda/q.$" We may therefore conclude from $(*)$ that

$$q = \Pr(X=1) \ge \Pr(Y=\lambda/q) = \Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).$$

If we choose $q = p(\lambda)/2,$ this says

$$p(\lambda)/2 \ge p(\lambda),$$

which is true only for non-positive numbers $p(\lambda).$ (You can check that this example still works when $\lambda=0,$ for then $Y$ reduces to the atom at $0$ and $\Pr(|Y|\gt |\lambda|/2) = \Pr(Y \ne 0) = 0.$) Therefore $0$ is the only universal lower bound, no matter what the value of $\lambda$ might be.

whuber
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