Does that sound about right?

Assume that $X$ is uniform on the interval $[0,1]$. Plot $P\big(\big|X\big|\geq \varepsilon\big)$ and $\dfrac{E\big(X^2\big)}{\varepsilon^2}$ as functions of $\varepsilon$. How does this relate to Chebyshev's inequality?


$ f(x) = \left\{ \begin{array}{l l} 1 & \quad for\,\,\, 0\leq x\leq 1\\ 0 & \quad \text{otherwise} \end{array} \right.$

\begin{eqnarray*} P(|x|\geq{\varepsilon})&=&\int_{\varepsilon}^1 1 \,\,\,dx\\ &=&x\Big |^1_{\epsilon}\\ &=&1-\varepsilon \,\,\,\,\,\,\,for\,\,\,\ 0\leq\varepsilon\leq 1\\\\ E(x^2)&=&\int^1_0x^2f(x)\,\,\,dx\\ &=&\dfrac{x^3}{3}\Big|^1_0=\dfrac{1}{3}\\\\ \dfrac{E(x^2)}{\varepsilon}&=&\dfrac{1}{3\varepsilon^2}\\ \end{eqnarray*}

for $\mu_u=0$:

\begin{eqnarray*} P(|x-0|\geq \varepsilon)&\leq& \dfrac{E(x-0)^2}{\varepsilon^2}\\ P(|x|\geq \varepsilon)&\leq &\dfrac{E(x)^2}{\varepsilon^2}\\ \end{eqnarray*}

I do struggle with the plot.

  • $\begingroup$ you struggle plotting $1-\epsilon$ and $\frac{1}{3\epsilon^2}$? wolframalpha.com/input/?i=1%2F%283x^2%29+and+1-x $\endgroup$ – Seyhmus Güngören Sep 12 '14 at 12:43
  • $\begingroup$ Wow! This is beautiful. Thank you so much. And in terms of the solution, does that look right? Is there anything else to improve? I would like to choose your feedback as the answer. $\endgroup$ – Olga Sep 12 '14 at 12:54

Yes your findings seem correct but you might want to see wiki for mean and variance etc. of a distribution of your interest. Here you have $k=1$ and $\mu=0$. BUT uniform distribution on $[0,1]$ has a mean $1/2$ not $0$. To have a zero mean uniform distribution you need to have it on $[-1/2,1/2]$.

Be careful with the last two lines. $E[X^2]\rightarrow$ first $X$ is capital it means it is a random variable and the term $(\cdot)^2$ should be inside the expectation.


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