<s>[INST] Using Chebyshev's inequality to obtain lower bounds Let $X_1$ and $X_2$ be i.i.d. continuous random variables with pdf $f(x) = 6x(1-x), 0<x<1$ and $0$, otherwise.
Using Chebyshev's inequality, find the lower bound of $P\left(|X_1 + X_2-1| \le\frac{1}{2}\right)$
What I did:
Using as Chebyshev's inequality,
$P(|X-\mu|\ge a)\le \frac{\sigma^2}{a^2}$
Where $a=\frac{1}{2}$
Finding the variance: $E[X^2] - (E[X])^2$
$  E[X] = \int_{0}^{1} x6x(1-x) dx$ $=6\left[\frac{x^  3}{3}-\frac{x^4}{4}\right]_0^1=6\left[\frac{1}{3}-\frac{1}{4}\right]=  \frac{6}{12}=\frac{1}{2}$
$  E[X^2]= \int_{0}^{1}x^{2}6x(1-x)dx=6\left[\frac{x^  4}{4}-\frac{x^5}{5}\right]_0^1=6\left[\frac{1}{4}-\frac{1}{5}\right]=\frac{6}{20}$
Therefore, $\sigma^2=\frac{6}{20}-\left(\frac{1}{2}\right)^2=\frac{1}{20}$
Putting in Chebyshev's inequality, 
$\frac{\sigma^2}{a^2} $= $\left[\frac{\frac{1}{20}}{\left(\frac{1}{2}\right)^2}\right]$=$\frac{4}{20}=\frac{1}{5} $
But what we need is $\le \frac{1}{2}$ which we get by $1-\frac{1}{5}=\frac{4}{5}$, 
But the answer is $\frac{3}{5}$
Where am I going wrong?
 [/INST] Comparing the given equation with Chebyshev's inequality, we get $\mu=1$
Since $X$ here has $X_1$ and $X_2$  i.i.d. continuous random variables, so both have same pdf and $E[X]=  E[X_1] + E[X_2]$
The mistake I did was to not calculate both $X_1$ and $X_2$ separately.
So the calculated $E[X]$ is actually$ E[X_1]=\frac{1}{2}$ and similarly $E[X_2]=\frac{1}{2} $
$E[X]=  E[X_1] + E[X_2]=\frac{1}{2}+ \frac{1}{2}=1$
And in the very same manner(recognising the same mistake everywhere) 
$E[X^2]=E[X_1^2] + E[X_2^2] = \frac{6}{20} + \frac{6}{20} = \frac{6}{10}$
Therefore variance $E[X^2] - (E[X])^2 =\frac{1}{10}$
Putting in Chebyshev's inequality, 
$\frac{\sigma^2}{a^2} $= $\left[\frac{\frac{1}{10}}{\left(\frac{1}{2}\right)^2}\right]$=$\frac{4}{10}=\frac{2}{5} $
So the lower bound which we get by $1-\frac{2}{5}=\frac{3}{5}$ which is the required answer.
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