4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

$$P(|X_1+X_2-1|>1/2)=P(X_1^2+X_2^2+2X_1X_2-2X_1-2X_2+1>1/4).$$

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

Thus

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.