MHB David's question from Yahoo Answers

[CaptainBlack]

Reposted from Yahoo Answers

2. Let X be a random variable that follows a Uniform(0; 1) distribution.
(a) Show that E(X) = 1/2 and Var(X) = 1/12.
(b) Using Chebyshev's inequality find an upper bound on the prob-
ability that X is more than k standard deviations away from its
expected value.
(c) Compute the exact probability that X is more than k standard
deviations from its expected value.
( d) Compare the bound to the exact probability.Thanks

 

[CaptainBlack]

(a) By definition of the expectation: \[E(X)= \int_{-\infty}^{\infty} x p(x)\; dx\]

but \(X \sim U(0,1)\) so \(p(x)=1\) for \(x\) in \([0,1]\) and zero otherwise this becomes:
\[E(X) = \int_0^1 x dx\]

Hence \[E(X)= \biggl[ \frac{x^2}{2} \biggr]_0^1= \frac{1^2}{2}-\frac{0^2}{2}=1/2 \]

Similarly for the variance:

\[V(X)=E( (X - \overline{X})^2 ) = \int_0^1 (x-1/2)^2\; dx = \frac{1}{12}\]

(b) I don't know what is being asked for here, Chebyshev's inequality is just this:
\[P( |X-\overline{X}| \ge k \sigma) \le \frac{1}{k^2}\]

(c) There are a number of ways of doing this, the easiest involves a diagram, but that is not convienient to use here, so we take the definition of the required probability:
\[P( |X-\overline{X}| \ge k \sigma)= P( X-\overline{X} \le - k \sigma) + P( X-\overline{X} \ge k \sigma)\]

Which may be written as a sum of integrals:
\[P( X-\overline{X} \le - k \sigma)=P(X \le \overline{X}-k \sigma)=\int_0^{\frac{1}{2}-\frac{k}{\sqrt{12}}} dx\] when \(k < \sqrt{12}/2\) and zero otherwise, and:

\[P( X-\overline{X} \ge - k \sigma)=P(X \ge \overline{X}+k \sigma)=\int_{\frac{1}{2}+\frac{k}{\sqrt{12}}}^1 dx\] when \(k < \sqrt{12}/2\) and zero otherwise.

CB