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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 1 fv(u) = /u³-¹e-³, u > 0

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 1 fv(u) = /u³-¹e-³, u > 0

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter2: Exponential, Logarithmic, And Trigonometric Functions
Section2.CR: Chapter 2 Review
Problem 111CR: Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data...
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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 k fv(u) = 1/2 u 2 e-²₂ u > 0 where 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 FT be the CDF of the random variable T. Verify that = 12 Z √U/k {1+P + P(Z² ≤ U) ¹P (Z² ≤ U) ift > 0, ift < 0. (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 ( 2² ≤ ²U) = [° F₂² (u) fv(u)du, < k where Fz2 is the CDF of Z2 and fu is the PDF of U.
Transcribed Image Text: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 k fv(u) = 1/2 u 2 e-²₂ u > 0 where 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 FT be the CDF of the random variable T. Verify that = 12 Z √U/k {1+P + P(Z² ≤ U) ¹P (Z² ≤ U) ift > 0, ift < 0. (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 ( 2² ≤ ²U) = [° F₂² (u) fv(u)du, < k where Fz2 is the CDF of Z2 and fu is the PDF of U.
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