-5
$\begingroup$

R.v. $Y_i$ is i.i.d. and normally distributed $N(0,\sigma^2)$ for all $i$. Prove that $$E\left(\frac{Y^2}{\sigma^2}\right) = 1$$ and $$W = \frac{1}{\sigma^2}\sum_{i=1}^n Y_i^2$$ is distributed $\chi_n^2$

Update:

As for $W \thicksim \chi^2(n)$, corollary to the Hint 2: If $Z_1,Z_2,...,Z_n$ are independent normal random variables with different means and variances, that is: $Z_i \thicksim N(\mu_i,\sigma_I^2)$ for $I = 1,2,...,n.$ Given that $W = \sum_{i=1}^{n} \frac{Y_i^2}{\sigma^2} = \sum_{i=1}^{n} Z_i^2$. Therefore, $W \thicksim X^2(n)$. I hope what I did is right.

$\endgroup$

closed as off-topic by Did, Juniven, Bram28, Shailesh, астон вілла олоф мэллбэрг Mar 15 '17 at 3:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Juniven, Bram28, Shailesh, астон вілла олоф мэллбэрг
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ How far did you get before posting this? What have you tried? $\endgroup$ – Sanderr Mar 14 '17 at 14:46
  • $\begingroup$ Can you determine the law of $Y_i/\sigma$? $\endgroup$ – Augustin Mar 14 '17 at 14:48
1
$\begingroup$

Hint 1: $\mathsf{Var}(Y_i) = \mathsf{E}(Y_i^2) - \mathsf{E}(Y_i)^2$

Hint 2: If $X_1, X_2, \cdots, X_k$ are i.i.d. standard normal random variables, then $\sum_i X_i^2 \sim \chi_k^2$.

$\endgroup$
  • $\begingroup$ How do I use the Hint 2? I used the Hint 1 but it got me nowhere...all I know is that $E(Y^2) = \sigma^2 + \mu^2$ $\endgroup$ – Zander Assand Mar 14 '17 at 16:23
  • $\begingroup$ standard normal r.v. is $Z = \frac{X-\mu}{\sigma}$ $\endgroup$ – Zander Assand Mar 14 '17 at 16:30
  • $\begingroup$ ok so $E(\frac{Y^2}{\sigma^2}) = E(\frac{Y}{\sigma})^2 = E(Z)^2$? $\endgroup$ – Zander Assand Mar 14 '17 at 16:38
  • $\begingroup$ In that case, $E(Z)^2 \thicksim \chi^2(1) = 1$ in which 1 stands for $r$ or degrees of freedom of 1 $\endgroup$ – Zander Assand Mar 14 '17 at 16:45
  • $\begingroup$ I cant see the use of Hint 2 from what I did. I only used a theorem that states that If $X \thicksim \chi^2(r)$, then $E[X] = r$. $\endgroup$ – Zander Assand Mar 14 '17 at 17:00

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