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So here i am getting E(X)=1/2,E(X^2)=0 and E(X^3)=Theta squared/4 How do i proceed now?How do i use the given x values ?

  • 2
    $\begingroup$ How do you obtain $E[X^2]=0$? That implies the variable must be constantly zero, which obviously is not the case. (The first expectation is incorrect, too.) You can sidestep some of these issues by considering how many moments you actually need to estimate this single parameter $\theta$. $\endgroup$
    – whuber
    Feb 6 '18 at 21:03
  • $\begingroup$ i took integration of x^2f(x) from -theta to theta and got 0 $\endgroup$
    – Abhisekkkk
    Feb 6 '18 at 21:10
  • $\begingroup$ Yes: that's what you wrote. But it's obviously wrong. $\endgroup$
    – whuber
    Feb 6 '18 at 21:11
  • 2
    $\begingroup$ You made an elementary mistake in your integration of $x^2$, it's not a question of how you should have approached it. As a hint: what is $- (-1)^3$? Furthermore, pay lots of attention to @whuber's last sentence in his first comment. $\endgroup$
    – jbowman
    Feb 6 '18 at 21:30
  • 2
    $\begingroup$ Using integration is okay but you should perform the integration correctly. Note that $x^2 f(x)$ is necessarily non-negative everywhere, so its integral cannot be 0 anywhere. You may be less likely to make an error if you use symmetry considerations to do the integral from 0 up to $\theta$. As with jbowman I urge you to consider how many moments you need. The fact that you calculated so many suggests that perhaps you don't know what the method of moments entails. If that's the case you should ask about that (but search first!). $\endgroup$
    – Glen_b
    Feb 6 '18 at 21:31

The first moment ($\mathbb{E}(X) =0$) by symmetry considerations. The second moment ($\mathbb{E}(X^2) = \frac{\theta^2}{3}$). Now equate this to the second sample moment ($\sum_{i=1}^4 x_i^2 /4 = \frac{2^2+1^2+\sqrt{5}^2+\sqrt{2}^2}{4} = 12/4=3$. So $\frac{\theta^2}{3}=3$ or $\theta = \pm3$ and we could then choose Answer (C).


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