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Questions about Linear Combinations of Random Variables

  • Thread starter ken15ken15
  • Start date
  • #1

Homework Statement


RQwRhjc.jpg



Homework Equations


Y=1/2*(X1-X3)^2+1/14*(X2+2X4-3X5)^2


The Attempt at a Solution


For (a) part, I have only learnt to find the moment-generating function of Y, but not finding the p.d.f.
Moreover, the examples I have seen only involves random variables Xi to the power 1, but not to higher power.

For (b) part, the difficulty for me is just similar to part (a).
 
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Answers and Replies

  • #2
761
13
Hint: [tex] (Z_1^2 + ... +Z_n^2) +(Z_1^2 + ... +Z_m^2) = (Z_1^2 + ... +Z_n^2 +...+ Z_m^2) [/tex]
 
  • #3
Hint: [tex] (Z_1^2 + ... +Z_n^2) +(Z_1^2 + ... +Z_m^2) = (Z_1^2 + ... +Z_n^2 +...+ Z_m^2) [/tex]
sorry but...what property is it? I haven't learnt this before...
 
  • #4
Ray Vickson
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Homework Statement


RQwRhjc.jpg



Homework Equations


Y=1/2*(X1-X3)^2+1/14*(X2+2X4-3X5)^2


The Attempt at a Solution


For (a) part, I have only learnt to find the moment-generating function of Y, but not finding the p.d.f.
Moreover, the examples I have seen only involves random variables Xi to the power 1, but not to higher power.

For (b) part, the difficulty for me is just similar to part (a).
##Z_1 = X_1 - X_3## has an ##N(0,a)## distribution, ##Z_2 = X_2+2X_4-3X_5## is ##N(0,b)##, and ##Z_1, Z_2## are independent. So ##Y## involves a weighted sum of squares of independent, mean-0 random variables.
 
  • #5
##Z_1 = X_1 - X_3## has an ##N(0,a)## distribution, ##Z_2 = X_2+2X_4-3X_5## is ##N(0,b)##, and ##Z_1, Z_2## are independent. So ##Y## involves a weighted sum of squares of independent, mean-0 random variables.
So do you mean given the random variables are independent, the sum or difference of them will also have the same distribution of them? Also, how can a duel with the square of the random variables? Do I need to consider it with transformation of variables?
 

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