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Let U x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be fv(u) = ut-¹e¯‡, u>0 where Z is the normalization constant making this function a PDF. Now, let Z ~ N(0, 1) be independent from U, and define T= Fr(t) Z √U/k Then Tt(k), a t distribution with k degrees of freedom. In this exercise, you will derive the form of the PDF of T following the same type of calculations in lecture. (i) Let Fr be the CDF of the random variable T. Verify that [+P (2² 0, - P(Z²

Let U x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be fv(u) = ut-¹e¯‡, u>0 where Z is the normalization constant making this function a PDF. Now, let Z ~ N(0, 1) be independent from U, and define T= Fr(t) Z √U/k Then Tt(k), a t distribution with k degrees of freedom. In this exercise, you will derive the form of the PDF of T following the same type of calculations in lecture. (i) Let Fr be the CDF of the random variable T. Verify that [+P (2² 0, - P(Z²

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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parts i and ii (asking iii, iv, v, in another question)

Exercise 6 Let U x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be fv(u) = /u-¹e, u>0 where Z is the normalization constant making this function a PDF. Now, let Z ~ N(0,1) be independent from U, and define Fr(t) = T = Then T t(k), a t distribution with k degrees of freedom. In this exercise, you will derive the form of the PDF of T following the same type of calculations in lecture. (i) Let Fr be the CDF of the random variable T. Verify that Z √U/k < [ + P(Z² < £U) - P(Z² ≤ U) (Hint: consider the event |Z| ≤ t√U/k and draw a picture of the PDF of Z.) (ii) Use the Law of Total Probability to show that P(2²sU)-F₂(x) fv(n)du, = where Fz² is the CDF of Z2 and fu is the PDF of U. (iii) Differentiate the integral to get that fr(t) = 1/2 √² fz² • (1/₁). fr(t) = constant x 2|t| k where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z² and U to get that ift > 0, ift < 0. -U. (Hint: Z²x²(1).) (v) Apply the change of variable v = · ( ¹ + ² ) u fu(u)du, √ ² u²+ - ¹ 6 - ² (1+² ) " du. U Jo u to get that fr(t) = constant x + (Hint: everything that doesn't involve t can be dumped into the constant factor.)
Transcribed Image Text:Exercise 6 Let U x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be fv(u) = /u-¹e, u>0 where Z is the normalization constant making this function a PDF. Now, let Z ~ N(0,1) be independent from U, and define Fr(t) = T = Then T t(k), a t distribution with k degrees of freedom. In this exercise, you will derive the form of the PDF of T following the same type of calculations in lecture. (i) Let Fr be the CDF of the random variable T. Verify that Z √U/k < [ + P(Z² < £U) - P(Z² ≤ U) (Hint: consider the event |Z| ≤ t√U/k and draw a picture of the PDF of Z.) (ii) Use the Law of Total Probability to show that P(2²sU)-F₂(x) fv(n)du, = where Fz² is the CDF of Z2 and fu is the PDF of U. (iii) Differentiate the integral to get that fr(t) = 1/2 √² fz² • (1/₁). fr(t) = constant x 2|t| k where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z² and U to get that ift > 0, ift < 0. -U. (Hint: Z²x²(1).) (v) Apply the change of variable v = · ( ¹ + ² ) u fu(u)du, √ ² u²+ - ¹ 6 - ² (1+² ) " du. U Jo u to get that fr(t) = constant x + (Hint: everything that doesn't involve t can be dumped into the constant factor.)
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