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?

  • $\begingroup$ What do you mean by "$X$"? Apparently it has the same distribution as $X_1$ and $X_2$, but how is that related to the question, which is about properties of the random variable $X_1+X_2$, which has a different distribution? $\endgroup$ – whuber Jan 24 '18 at 16:57
  • $\begingroup$ Why does it have a different distribution when $X_1$ and $X_2$ are i.i.d random variables? I am sorry I might have misunderstood the question. Tell me how to go about it? I am fairly new to this concept $\endgroup$ – Shreya Bhandari Jan 24 '18 at 17:01
  • $\begingroup$ Just because $X_1$ and $X_2$ have the same distribution as each other does not mean that $|X_1 + X_2 -1|$ has the same distribution as the two of them. $\endgroup$ – jbowman Jan 24 '18 at 17:28
  • $\begingroup$ Okay, so how would you know the pdf there then? Also, what i just noticed is that my calculated $E(X)$ is $ \frac{1}{2}$ but if we match with the Chebyshev's inequality, $\mu=1$, so if we do notice $X_1 $and$ X_2$ being i.i.d $E(X_1+X_2)=E(X_1)+E(X_2) = \frac{1}{2}+\frac{1}{2}$, so adjusting that way(multiplying by 2 everywhere) we would get $\sigma^2 =\frac{2}{20}$ which would give us the required answer $\frac{3}{5}$. Does that make any sense? $\endgroup$ – Shreya Bhandari Jan 24 '18 at 17:48
  • $\begingroup$ Shreya, you made a key point: you don't need to know the distribution of $X_1+X_2$; you only need to know its mean and variance. And indeed its mean is twice the mean of either $X_i$ and its variance is twice the variance of either $X_i$. $\endgroup$ – whuber Jan 24 '18 at 23:04

You dropped a 6. It should be $E[X^2]=6/20=3/10$.

$$P(|X_1+X_2-1|\leq 1/2)=1-P(|X_1+X_2-1|>1/2).$$


$$P(|X_1+X_2-1|^2>1/4)\leq \frac{6/10+2(1/2)^2-1-1+1}{1/4}=4/10.$$


$$P(|X_1+X_2-1|\leq 1/2)\geq 1-4/10=3/5$$

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  • $\begingroup$ Hey, that's a typo, I'll edit that, you can see that I've taken $\frac{6}{20}$ only while calculating $\sigma^2$. Also, could you explain the $\frac{6/10+2(1/2)^2-1-1+1}{1/4}$ part? $\endgroup$ – Shreya Bhandari Jan 24 '18 at 19:17
  • $\begingroup$ $P(Y\geq \epsilon) \leq E[Y]/\epsilon$ $\endgroup$ – Alex R. Jan 24 '18 at 20:15
  • $\begingroup$ But you put the value of $P(|X_1+X_2-1|^2>1/4)$ ie $4/10$ in the formula where the value of $P(|X_1+X_2-1|>1/2)$ was needed. Shouldn't the square root of $4/10$ be used instead? $\endgroup$ – Shreya Bhandari Jan 24 '18 at 21:25
  • $\begingroup$ I squared both sides within the probability prior to applying the identity. $\endgroup$ – Alex R. Jan 24 '18 at 21:45
  • $\begingroup$ What formula are you applying to produce your "$\le$"? On the face of it you have replaced each variable in the event $X_1^2 + \cdots -2X_2+1 \gt 1/4$ with its expectation divided by $1/4$, but that makes no sense at all. $\endgroup$ – whuber Jan 24 '18 at 23:22

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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