Answer to Question #338298 in Statistics and Probability for peace boy

Question #338298
  1. Let X1, X2 , and X3 be independent standard normal random variable. If we defined Y1=X2, Y2=X1+X2/2 and Y3=X1+ X2 +X3/3.Then find then joint pdf of Y1, Y2, and Y3 using the Jacobian method?
1
Expert's answer
2022-05-12T14:43:19-0400

Suppose that random variables X1X_1, X2X_2, X3X_3 have normal distributions with parameters μ1,σ12\mu_1,\sigma^2_1, μ2,σ22\mu_2,\sigma^2_2 and μ3,σ32\mu_3,\sigma^2_3 respectively. We use a known fact that in case a random variable ZZ has a normal distribution with parameters μZ,σZ2\mu_Z,\sigma^2_Z, then, a random variable aZaZ, where aR,a\in{\mathbb{R}}, has a normal distribution with parameters aμZ,a2σZ2a\mu_Z,a^2\sigma^2_Z. We receive, that X22\frac{X_2}{2} has a normal distribution with parameters μ22,σ224\frac{\mu_2}{2},\frac{\sigma^2_2}{4} and X33\frac{X_3}{3} has a normal distribution with parameters: μ33,σ329\frac{\mu_3}{3},\frac{\sigma^2_3}{9}. We use also a known fact, that a sum of normally distributed variables is again a normally distributed variable. We get, that Y2Y_2 has a normal distribution with parameters: μ1+μ22,σ12+σ224\mu_1+\frac{\mu_2}2,\sigma_1^2+\frac{\sigma_2^2}{4} and Y3Y_3 has a normal distribution with parameters: μ1+μ2+μ33,σ12+σ22+σ329\mu_1+\mu_2+\frac{\mu_3}3,\sigma_1^2+\sigma_2^2+\frac{\sigma_3^2}{9}. Denote: α1=μ1+μ22\alpha_1=\mu_1+\frac{\mu_2}2, β1=σ12+σ224\beta_1=\sqrt{\sigma_1^2+\frac{\sigma_2^2}{4}}, α2=μ1+μ2+μ33\alpha_2=\mu_1+\mu_2+\frac{\mu_3}3, β2=σ12+σ22+σ329\beta_2=\sqrt{\sigma_1^2+\sigma_2^2+\frac{\sigma_3^2}{9}}. We use a known fact that the joint probability density function is equal to the product of density functions for independent variables. We get: pY1,Y2,Y3(y1,y2,y3)=pY1(y1)pY2(y2)pY3(y3)=1σ2β1β2(2π)3e12(y1μ2σ2)212(y2α1β1)212(y3α2β2)2p_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)=p_{Y_1}(y_1)p_{Y_2}(y_2)p_{Y_3}(y_3)=\frac{1}{\sigma_2\beta_1\beta_2(\sqrt{2\pi})^3}e^{-\frac12(\frac{y_1-\mu_2}{\sigma_2})^2-\frac12(\frac{y_2-\alpha_1}{\beta_1})^2-\frac12(\frac{y_3-\alpha_2}{\beta_2})^2}

Answer: pY1,Y2,Y3(y1,y2,y3)=1σ2β1β2(2π)3e12(y1μ2σ2)212(y2α1β1)212(y3α2β2)2.p_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)=\frac{1}{\sigma_2\beta_1\beta_2(\sqrt{2\pi})^3}e^{-\frac12(\frac{y_1-\mu_2}{\sigma_2})^2-\frac12(\frac{y_2-\alpha_1}{\beta_1})^2-\frac12(\frac{y_3-\alpha_2}{\beta_2})^2}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment