# Thread: [SOLVED] Finding distribution

1. ## [SOLVED] Finding distribution

Let $\displaystyle Y_1, Y_2, ..., Y_5$ be a random sample of size 5 from a normal distribution with mean 0 and variance 1, and let

$\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$

Let $\displaystyle Y_6$ be another independent observation from the same distribution

a) What is the distribution of $\displaystyle W=\sum\limits_{i=1}^5 Y_i^2$? I think it's $\displaystyle \chi_5$
b) What is the distribution of $\displaystyle U= \sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$?

c) What is the distribution of $\displaystyle \sum\limits_{i=1}^5(Y_i-\overline{Y})^2 + Y_6^2$?

2. Originally Posted by noob mathematician
Let $\displaystyle Y_1, Y_2, ..., Y_5$ be a random sample of size 5 from a normal distribution with mean 0 and variance 1, and let

$\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$

Let $\displaystyle Y_6$ be another independent observation from the same distribution

a) What is the distribution of $\displaystyle W=\sum\limits_{i=1}^5 Y_i^2$? I think it's $\displaystyle \chi_5$
b) What is the distribution of $\displaystyle U= \sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$?

c) What is the distribution of $\displaystyle \sum\limits_{i=1}^5(Y_i-\overline{Y})^2 + Y_6^2$?
You will be flattered; there's so much information given you in the first two lines. It's a give away for you to earn easy scores.

Hint: the shape of you distribution is a perfect symmety.

3. Originally Posted by novice
You will be flattered; there's so much information given you in the first two lines. It's a give away for you to earn easy scores.

Hint: the shape of you distribution is a perfect symmety.
Pardon me if i don't get your hint: so the answer is $\displaystyle \chi_5, \chi_5, \chi_6$?

4. Originally Posted by noob mathematician
Pardon me if i don't get your hint: so the answer is $\displaystyle \chi_5, \chi_5, \chi_6$?
That is the correct distribution followed by W (I think novice misread the question ....)

5. Originally Posted by mr fantastic
That is the correct distribution followed by W (I think novice misread the question ....)
Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

b)Find the distribution of $\displaystyle U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$.

I actually expanded the question:

since $\displaystyle \sum\limits_{i=1}^n (X_i-\overline{X})^2$
=$\displaystyle \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-n\overline{X}^2$

Applying to my question will yield:

$\displaystyle \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2$,
then since it was defined initially that $\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$, then it should have a distribution of N(0,1) too.

Subsequently, I will have: $\displaystyle U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2$, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as $\displaystyle {\chi_5}^2$?

6. Originally Posted by noob mathematician
Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

b)Find the distribution of $\displaystyle U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$.

I actually expanded the question:

since $\displaystyle \sum\limits_{i=1}^n (X_i-\overline{X})^2$
=$\displaystyle \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-n\overline{X}^2$

Applying to my question will yield:

$\displaystyle \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2$,
then since it was defined initially that $\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$, then it should have a distribution of N(0,1) too.

Subsequently, I will have: $\displaystyle U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2$, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as $\displaystyle {\chi_5}^2$?
Please accept my sincerely apology. Indeed I misread the question as Mr. Fantastic said.
Now it is clear that the question is more interesting than I originally thought. Admittedly, I my self have something to gain by solving it, but I doubt that I could beat you to it.

7. Originally Posted by noob mathematician
Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

b)Find the distribution of $\displaystyle U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$.

I actually expanded the question:

since $\displaystyle \sum\limits_{i=1}^n (X_i-\overline{X})^2$
=$\displaystyle \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2$
=$\displaystyle \sum\limits_{i=1}^nX_i^2-n\overline{X}^2$

Applying to my question will yield:

$\displaystyle \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2$,
then since it was defined initially that $\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$, then it should have a distribution of N(0,1) too.

Subsequently, I will have: $\displaystyle U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2$, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as $\displaystyle {\chi_5}^2$?
So any view on my above comment?

8. a) is $\displaystyle \chi^2_5$

next use $\displaystyle {(n-1)S^2\over\sigma^2}\sim\chi^2_{n-1}$

with $\displaystyle \sigma^2=1$

we have in (b) $\displaystyle {(n-1)S^2\over\sigma^2}\sim\chi^2_4$

hence (c) is $\displaystyle \chi^2_5$

since $\displaystyle \chi^2_4 + \chi^2_1=\chi^2_5$

using independence, via the MGF technique

9. Originally Posted by matheagle
a) is $\displaystyle \chi^2_5$

next use $\displaystyle {(n-1)S^2\over\sigma^2}\sim\chi^2_{n-1}$

with $\displaystyle \sigma^2=1$

we have in (b) $\displaystyle {(n-1)S^2\over\sigma^2}\sim\chi^2_4$

hence (c) is $\displaystyle \chi^2_5$

since $\displaystyle \chi^2_4 + \chi^2_1=\chi^2_5$

via independence, via the MGF technique
Thanks, it's a cool solution.

So now I'm interested to know the next distribution:

$\displaystyle 2(5\overline{Y}^2+Y_6^2)/\chi_4^2$ It seems like it's a F-distribution, but I can't seem to extract the equation to form pattern like $\displaystyle F_{m,n}\rightarrow (U/m)/(V/n)$

Btw, just curious that for the nominator: $\displaystyle (5\overline{Y}^2+Y_6^2)$, does it convert to $\displaystyle \chi_6^2$? Then does that mean that $\displaystyle 2(5\overline{Y}^2+Y_6^2)$ will be $\displaystyle \chi_{12}^2$??

10. no
and you must have that chi-square independent of y bar and y6.

It's an F2,4

$\displaystyle Y_i\sim N(0,1)$

so $\displaystyle \bar Y\sim N(0,1/5)$

or $\displaystyle \sqrt{5}\bar Y\sim N(0,1)$

Thus $\displaystyle 5\bar Y^2\sim \chi^2_1$

Since $\displaystyle Y_6^2\sim \chi^2_1$ we have

$\displaystyle 5\bar Y^2 +Y_6^2\sim \chi^2_2$

HENCE $\displaystyle {{5\bar Y^2 +Y_6^2\over 2}\over {\chi^2_4\over 4}}\sim F_{2,4}$

The 2 in the numerator is just algebra, 4/2.

11. Originally Posted by matheagle
no
and you must have that chi-square independent of y bar and y6.

It's an F2,4

$\displaystyle Y_i\sim N(0,1)$

so $\displaystyle \bar Y\sim N(0,1/5)$

or $\displaystyle \sqrt{5}\bar Y\sim N(0,1)$

Thus $\displaystyle 5\bar Y^2\sim \chi^2_1$
Hi thanks again, I'm sure you are right.

However, I was kinda confused with the above transformation, such as how it jumps from $\displaystyle \bar Y\sim N(0,1/5)$ to $\displaystyle \sqrt{5}\bar Y\sim N(0,1)$

12. you obtain a N(0,1) by subtracting off the mean and dividing by the st deviation

$\displaystyle Z={W-\mu\over \sigma}={\bar Y-0\over 1/{\sqrt 5}}={\sqrt 5}\bar Y\sim N(0,1)$