variance = Sum Xsq/n - (Xav.) Sq = Sum ( Xi - Xav.)Sq
Expanding you will get this.
Alright this is a question from my stats class, I'm sure most will find it basic , however im not sure if i'm on the right track or completely off.
Show that: ∑i-n (Xi-Xbar)^2 = ∑i-n Xi^2-nXbar^2
Would I use the formula for the sample mean to then prove this, or is that not right.
Any help would be appreciated .
p.s. sorry for the horrible notation, i'm not exactly sure where to find the proper symbols.
variance = Sum Xsq/n - (Xav.) Sq = Sum ( Xi - Xav.)Sq
Expanding you will get this.
Great! thanks for the replies.
So then would I have?:
n Xbar = ∑i-n (Xi)
nXbar^2=∑i-n (Xi)^2
0=∑i-n (Xi)^2-n Xbar^2
=∑i-n (Xi-nXbar)^2
To be honest I dont know, I figured in order to make it equivalent to the first equation that would be the way to go. Any suggestions?
Great Thanks! I had started off simply expanding it as you orginally said (a-b)= a-b)^2 = a^2 -2ab +b^2 however I didnt think of using the sample mean equation, so again thanks for pointing that out
Last edited by canadia; 09-20-2009 at 08:17 AM.
If you wanted to remember this for long time. Try to solve this using some example.. this will give more insight of the calculation.
ex: X1=5, X2 =2 ,X2 =1 here n = 3
In the long run, we're all dead.
If It isnt too much trouble, I had one more question.
Consider a transformed variable Wi, where Wi=a+bXi. Calculate the sample mean and variance of Wi, in terms of the sample mean and variance of Xi.
So in terms of Xi, we would have Xi=Wi-a/b
So if we take the sample mean formula Xbar= ∑i-n Xi/n, and replace Xi with Wi, is that what the question is asking for or is it something completely different.?
Good idea, I'm going to do that.If you wanted to remember this for long time. Try to solve this using some example.. this will give more insight of the calculation.
ex: X1=5, X2 =2 ,X2 =1 here n = 3
You may have to bit careful on your algebra.. Does Wi=a+bXi. implies
Xi=Wi-a/b ? or Xi=(Wi-a)/b?
Hint:
Sample mean of Wi = 1/n * ∑i-n Wi
Now replace Wi in terms of Xi and finally you will get answer in terms of Xbar..
Similar expression for the Variance. of Wi
In the long run, we're all dead.
∑i-n(Xi-Xbar)^2 = ∑i-n { Xi^2 -2Xi Xbar + Xbar^2 )
= ∑i-n Xi^2 - 2 Xbar ∑i-n Xi + ∑i-n Xbar^2
= ∑i-n Xi^2 - 2 Xbar [ nXbar] + nXbar^2
=∑i-n Xi^2 - nXbar^2
Sorry I just had one more question about this, I understand every transformation except in step one we had ∑i-n Xbar^2 and in step 2 this became nXbar^2, could you clarify this for me.
Im sure this is annoying but I would appreciate it .
∑i-n Xbar^2 = { Xbar^2 + Xbar^2+ ... (n times )+Xbar^2 }
= n Xbar^2
{ remember algebra .. K + K+ K = 3K }
In the long run, we're all dead.
Vinux, you have been a great help thank you. One more thing, to you what does d= 1/n* ∑i-n (Xi-Xbar) equal to? Personally I was thinking it would be the mean deviation?
If there is absolute function.. then you are right.. MD
Mean Deviation = 1/n* ∑i-n |(Xi-Xbar)| or 1/n* ∑i-n ABS(Xi-Xbar)
Otherwise 1/n* ∑i-n (Xi-Xbar) will be zero..
In the long run, we're all dead.
Vinux, I have another question for you if you dont mind.
Suppose {Aj}j-m is a collection of mutally exclusive, collectively exhaustive events in the sample space, and let B an event with P(B) >0. Show that ∑j-m P(Aj|B)=1.
Alright so I'm thinking P(Aj|B)= P(Aj∩ B)/P(B)
And since they are mutually exclusive A∩ B= 0?
Also, since they are collectively exhaustive AUB=S=1?
He A1, A2, A3 ...Am make the partition. Aj's are mutually exclusive.
But Aj's and B are not exclusive..
See the diagram below
In this E1 and E2 you can consider as Ajs . And E is B
In the long run, we're all dead.
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