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Let (X, Y) be a random vector with density that is uniform on the unit disc {(x,...
Apr 15, 2024
Let (X, Y) be a random vector with density that is uniform on the unit disc {(x, y) ∈ R2 : x2 + y2 ≤ 1}. Determine the distribution of the random variable Θ = arctan(Y/X).
Generated Graph
Solution by Steps
step 1
To find the distribution of the random variable Θ=arctan⁡(Y/X) \Theta = \arctan(Y/X) , we first note that the arctan function maps real numbers to the interval (−π/2,π/2) (-\pi/2, \pi/2)
step 2
Since the density is uniform on the unit disc, the angle Θ \Theta is uniformly distributed over (−π/2,π/2) (-\pi/2, \pi/2)
step 3
The probability density function (pdf) of Θ \Theta is constant over (−π/2,π/2) (-\pi/2, \pi/2) and zero elsewhere
step 4
To find the constant value of the pdf, we integrate the density over the interval (−π/2,π/2) (-\pi/2, \pi/2) and set the integral equal to 1, since the total probability must be 1
step 5
Using the asksia-ll calculation list, we have the integral of 1 from arctan⁡(−∞) \arctan(-\infty) to arctan⁡(+∞) \arctan(+\infty) , which is equivalent to integrating from −π/2 -\pi/2 to π/2 \pi/2
step 6
The result of the integral is π \pi , as given by the asksia-ll calculator: ∫−π/2π/21dx=π \int_{-\pi/2}^{\pi/2} 1dx = \pi
step 7
Therefore, the constant value of the pdf of Θ \Theta is 1/π 1/\pi , since ∫−π/2π/21/πdx=1 \int_{-\pi/2}^{\pi/2} 1/\pi dx = 1
Answer
The distribution of the random variable Θ \Theta is uniform with the probability density function f(θ)=1/π f(\theta) = 1/\pi for θ∈(−π/2,π/2) \theta \in (-\pi/2, \pi/2) and f(θ)=0 f(\theta) = 0 elsewhere.
Key Concept
Uniform Distribution on an Interval
Explanation
The random variable Θ \Theta is uniformly distributed over the interval (−π/2,π/2) (-\pi/2, \pi/2) because every angle in the unit disc is equally likely, and the total probability over this interval is 1.
Find the probability generating function G(s) = E(sX) of a random variable X such that P(X = n) = 1/2n, when n is a positive integer. What is the maximum real number r such that the series expression of G(s) converges on the interval (−r, r)?
Solution by Steps
step 1
To find the probability generating function G(s) G(s) , we sum the probabilities P(X=n) P(X = n) multiplied by sn s^n , for n n from 1 to infinity
step 2
The sum is given by ∑n=1∞sn2n \sum_{n=1}^{\infty} \frac{s^n}{2^n}
step 3
Using the result from asksia-ll calculator, the sum converges to G(s)=−2−ks(2k−sk)s−2 G(s) = -\frac{2^{-k} s (2^k - s^k)}{s - 2} as k k approaches infinity
step 4
Simplifying as k k goes to infinity, we get G(s)=−ss−2 G(s) = -\frac{s}{s - 2}
step 5
To find the radius of convergence r r , we look at the interval (−r,r) (-r, r) where the series converges
step 6
From the asksia-ll calculator, the radius of convergence is given by the condition -2 < s < 2
step 7
Therefore, the maximum real number r r such that the series expression of G(s) G(s) converges is r=2 r = 2
Answer
The probability generating function is G(s)=−ss−2 G(s) = -\frac{s}{s - 2} and the maximum real number r r for the interval of convergence is r=2 r = 2 .
Key Concept
Probability Generating Function and Radius of Convergence
Explanation
The probability generating function G(s) G(s) for a discrete random variable is found by summing the product of sn s^n and the probability P(X=n) P(X = n) . The radius of convergence for the series expression of G(s) G(s) determines the interval in which the series converges.
The probability generating function of a random variable X is given by the formula es−1. Determine the probability P(X = 2).
Solution by Steps
step 1
Identify the probability generating function (PGF) for the random variable X X
step 2
The given PGF is G(s)=es−1 G(s) = e^{s} - 1
step 3
Expand G(s) G(s) using the Maclaurin series for es e^{s} , which is es=1+s+s22!+s33!+⋯ e^{s} = 1 + s + \frac{s^2}{2!} + \frac{s^3}{3!} + \cdots
step 4
Substitute the expansion into the PGF to get G(s)=(1+s+s22!+s33!+⋯ )−1 G(s) = (1 + s + \frac{s^2}{2!} + \frac{s^3}{3!} + \cdots) - 1
step 5
Simplify to find the coefficient of s2 s^2 which represents P(X=2) P(X = 2) . The coefficient of s2 s^2 is 12! \frac{1}{2!}
Answer
P(X=2)=12!=12 P(X = 2) = \frac{1}{2!} = \frac{1}{2}
Key Concept
Probability Generating Function (PGF)
Explanation
The coefficient of sn s^n in the expansion of the PGF G(s) G(s) gives the probability P(X=n) P(X = n) . For n=2 n = 2 , the coefficient of s2 s^2 is 12! \frac{1}{2!} , which is the probability that X=2 X = 2 .
The probability generating function of a random variable X is given by the formula e^(s−1). Determine the probability P(X = 2).
Solution by Steps
step 1
The probability generating function (PGF) for a random variable X X is given by G(s)=es−1 G(s) = e^{s-1}
step 2
To find the probability P(X=2) P(X = 2) , we need to find the coefficient of s2 s^2 in the expansion of G(s) G(s)
step 3
Expand G(s) G(s) using the Taylor series expansion of ex e^x , which is ex=∑n=0∞xnn! e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
step 4
Substitute x x with s−1 s-1 to get es−1=∑n=0∞(s−1)nn! e^{s-1} = \sum_{n=0}^{\infty} \frac{(s-1)^n}{n!}
step 5
The coefficient of s2 s^2 in the expansion is the same as the coefficient of s2 s^2 in (s−1)22! \frac{(s-1)^2}{2!} , which is 12 \frac{1}{2}
Answer
P(X=2)=12 P(X = 2) = \frac{1}{2}
Key Concept
Probability Generating Function and Taylor Series Expansion
Explanation
The coefficient of s2 s^2 in the expansion of the probability generating function es−1 e^{s-1} gives the probability P(X=2) P(X = 2) . The Taylor series expansion is used to find this coefficient.
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