A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
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8.1
Q 8.1. Suppose that X and Y are two random variables whose joint distribution is the standard Gaussian distribution in the plane. Let random variables R≥ 0 and = = [0, 2π) satisfy X = R cos and Y Rsin 0. (a) State the density of the joint distribution of R and O. (You do not need to derive this) (b) Express Z = in terms of and hence, making reference to part (a), calculate the density of the distribution of Z. It will help to try to picture this one.
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Transcribed Image Text:Q 8.1. Suppose that X and Y are two random variables whose joint distribution is the standard Gaussian distribution in the plane. Let random variables R≥ 0 and = = [0, 2π) satisfy X = R cos and Y Rsin 0. (a) State the density of the joint distribution of R and O. (You do not need to derive this) (b) Express Z = in terms of and hence, making reference to part (a), calculate the density of the distribution of Z. It will help to try to picture this one.
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Step 1: Write the given information
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Step 2: State the density of the joint distribution of R and ϴ
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Step 3: Calculate the density of the distribution of Z=Y/X in terms of ϴ
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for part b, could u please provide more details of the calculations I circled? thanks

Step3 c) (b) Let the random variable Z = F₂(2)=P(X ≤2) Put = P(Y ≤ ZX, X ≥0) + P(Y ≤ ZX, X <0) = 2P(Y ≤ ZX, X>0) XZ = 25° 5x² fx(x) fy(v) dy dx 0 -∞ V 2 2 = ²5² ²2/12 - 0 + 1 = 0 ²+² oy ox 2√² √x² e e dy dx 0 √√2π √2π ∞ XZ = 2 × ²2²1 ²²² ²²0 +² oy dx 2 2 e e dy 2₁ 0 -∞ XZ -15 oyce why 1.00 disappears? -22 = e dy dx π -∞ details? = 1/1 1 0 0 e 0 2 Differentiate the function to x² 2 (xz)² How to diffe √₂ (2) = = √²0 = = = (42²³² 1 2 e e π 0 Ell > × e Y X ?Cound the pdf of Z, give и more - 1/²/(1+2²) x dx x dx
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Transcribed Image Text:Step3 c) (b) Let the random variable Z = F₂(2)=P(X ≤2) Put = P(Y ≤ ZX, X ≥0) + P(Y ≤ ZX, X <0) = 2P(Y ≤ ZX, X>0) XZ = 25° 5x² fx(x) fy(v) dy dx 0 -∞ V 2 2 = ²5² ²2/12 - 0 + 1 = 0 ²+² oy ox 2√² √x² e e dy dx 0 √√2π √2π ∞ XZ = 2 × ²2²1 ²²² ²²0 +² oy dx 2 2 e e dy 2₁ 0 -∞ XZ -15 oyce why 1.00 disappears? -22 = e dy dx π -∞ details? = 1/1 1 0 0 e 0 2 Differentiate the function to x² 2 (xz)² How to diffe √₂ (2) = = √²0 = = = (42²³² 1 2 e e π 0 Ell > × e Y X ?Cound the pdf of Z, give и more - 1/²/(1+2²) x dx x dx
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for part b, could u please provide more details of the calculations I circled? thanks

Step3 c) (b) Let the random variable Z = F₂(2)=P(X ≤2) Put = P(Y ≤ ZX, X ≥0) + P(Y ≤ ZX, X <0) = 2P(Y ≤ ZX, X>0) XZ = 25° 5x² fx(x) fy(v) dy dx 0 -∞ V 2 2 = ²5² ²2/12 - 0 + 1 = 0 ²+² oy ox 2√² √x² e e dy dx 0 √√2π √2π ∞ XZ = 2 × ²2²1 ²²² ²²0 +² oy dx 2 2 e e dy 2₁ 0 -∞ XZ -15 oyce why 1.00 disappears? -22 = e dy dx π -∞ details? = 1/1 1 0 0 e 0 2 Differentiate the function to x² 2 (xz)² How to diffe √₂ (2) = = √²0 = = = (42²³² 1 2 e e π 0 Ell > × e Y X ?Cound the pdf of Z, give и more - 1/²/(1+2²) x dx x dx
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Transcribed Image Text:Step3 c) (b) Let the random variable Z = F₂(2)=P(X ≤2) Put = P(Y ≤ ZX, X ≥0) + P(Y ≤ ZX, X <0) = 2P(Y ≤ ZX, X>0) XZ = 25° 5x² fx(x) fy(v) dy dx 0 -∞ V 2 2 = ²5² ²2/12 - 0 + 1 = 0 ²+² oy ox 2√² √x² e e dy dx 0 √√2π √2π ∞ XZ = 2 × ²2²1 ²²² ²²0 +² oy dx 2 2 e e dy 2₁ 0 -∞ XZ -15 oyce why 1.00 disappears? -22 = e dy dx π -∞ details? = 1/1 1 0 0 e 0 2 Differentiate the function to x² 2 (xz)² How to diffe √₂ (2) = = √²0 = = = (42²³² 1 2 e e π 0 Ell > × e Y X ?Cound the pdf of Z, give и more - 1/²/(1+2²) x dx x dx
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