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# Random Variables

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1. (a) Show that if Z1 and Z2 are independent standard normal random variables, then for all &#961; (correlation), Z1 and &#961;Z1+sqrt(1-&#961;2)*Z2 are standard normal with correlation &#961;.

(b) Show that for all &#961; and v, T1 = (Z1)/sqrt(X/v)and T2 = (Z2)/sqrt(X/v)
have correlation &#961;, where Z1and Z2 are standard normal with correlation &#961;, and X is independent of both Z1 and Z2 and has a Chi-Squared distribution. (for simplification use the fact that T1 and T2 both have the t distribution with v degrees of freedom and conditional expectations.)
(v=degrees of freedom)

##### Solution Summary

1. (a) Show that if Z1 and Z2 are independent standard normal random variables, then for all &#961; (correlation), Z1 and &#961;Z1+sqrt(1-&#961;2)*Z2 are standard normal with correlation &#961;.

(b) Show that for all &#961; and v, T1 = (Z1)/sqrt(X/v)and T2 = (Z2)/sqrt(X/v)
have correlation &#961;, where Z1and Z2 are standard normal with correlation &#961;, and X is independent of both Z1 and Z2 and has a Chi-Squared distribution. (for simplification use the fact that T1 and T2 both have the t distribution with v degrees of freedom and conditional expectations.)
(v=degrees of freedom)

##### Solution Preview

1. (a) Show that if Z1 and Z2 are independent standard normal random variables, then for all ρ (correlation), Z1 and ρZ1+sqrt(1-ρ2)*Z2 are standard normal with correlation ρ.
Proof. Since Z1 and Z2 are independent standard normal random variables, we have
E(Z1)=E(Z2)=0, Var(Z1)=Var(Z2)=1 , E(Z1Z2)=E(Z1)E(Z2)=0.
We know that . Similarly, . So,

We now try to find

So, the correlation coefficient of Z1 and is

(b) Show that for all ρ and v, T1 = (Z1)/sqrt(X/v)and T2 = (Z2)/sqrt(X/v)
have correlation ρ, where Z1and Z2 ...

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###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"

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