Question

X N(0,1) and Z 3X2 + 1, what is the PDF of Z'?

Answer 1

Solution

Back-up Theory

If X ~ N(0, 1), then Y = X² ~ χ²₁ i.e., Chi-square distribution with 1 degree of freedom ............ (1)

If X ~ χ²_ν i.e., Chi-square distribution with ν degrees of freedom, its pdf is:

f(x: ν) = {x^{(ν/2) – 1}.e^-(x/2)}/{2^ν/2.ℾ(ν/2)}, x > 0, where ℾ is the gamma function, ℾn = (n - 1)!...... (2)

In particular, if X ~ χ²₁, the pdf is: f(x: 1) = {x^{(1/2) – 1}.e^-(x/2)}/{2^1/2.√π}, x > 0 ........................... (2a)

(1) and (2a) => pdf of Y is: f(y: 1) = {y^{(1/2) – 1}.e^-(y/2)}/{2^1/2.√π}, x > 0 ..................................... (3)

Now to work out the solution,

Given X ~ N(0, 1) and Z = 3X² + 1, Z = 3Y + 1, where Y = X² ………………………………… (4)

Now, Z = z => 3Y + 1 = z or Y = (z – 1)/3. Hence,

f_Z(z) = f_Y[(2/3)z]

= [{(z – 1)/3 }^{(1/2) – 1}.e^{-(z – 1)/6)}/{2^1/2.√π}] Answer [Vide (1) and (2a)]

DONE