MHB A question about 'properties of multivariate normal distributions'.

rof
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hello guys,
i have a question and looking for an answer quickly...

question;

prove that,
XxZdUxc.png

Z is a vector.thank you.
 
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rof said:
hello guys,
i have a question and looking for an answer quickly...

question;

prove that,

Z is a vector.thank you.

Hi rof! Welcome to MHB! ;)

The chi-square distribution is defined as the sum of the squares of p independent standard normal distributions.
So your proof follows from the definition.
 
this is the full question paper. (if that helps you to understand the nature of the question)
the answer i want is to question number 5.

View attachment 5906

thank you
 

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rof said:
this is the full question paper. (if that helps you to understand the nature of the question)
the answer i want is to question number 5.http://i.imgur.com/l2CpJz3.jpgthank you

That link is broken...and rather than using a free online image hosting service (many of which don't work well), it is better to attach images inline so that everyone can see them and they are not subject to the image hosting service taking them down.
 
MarkFL said:
That link is broken...and rather than using a free online image hosting service (many of which don't work well), it is better to attach images inline so that everyone can see them and they are not subject to the image hosting service taking them down.

i corrected it by editing the post as soon as i posted it, but somehow editing didnt happen.

here is the link;
http://i.imgur.com/QnBxP5g.jpg

5th question
 
rof said:
i corrected it by editing the post as soon as i posted it, but somehow editing didnt happen.

here is the link;
http://i.imgur.com/QnBxP5g.jpg

5th question

I went ahead and edited your previous post to attach the image inline. (Yes)
 
rof said:
this is the full question paper. (if that helps you to understand the nature of the question)
the answer i want is to question number 5.

thank you

Your vector $\mathbf Z$ consists of $p$ independent standard normal distributions.
When we calculate the dot product $\mathbf Z' \mathbf Z$, we get the sum of the squares of those $p$ independent standard normal distributions.
That is a $\chi_p^2$ distribution by definition.
 
thanks all for taking time to read my question and providing answers.
good luck to all... :)
 
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