Multiple Choice Given that the intercepts of a graph are ( − 4 , 0 ) and ( 0 , 5 ) , choose the statement that is true The y -intercept is − 4 , and the x -intercept is 5 . The y -intercepts are − 4 and 5 . The x -intercepts are − 4 and 5 . The x -intercept is − 4 , and the y -intercept is 5 .
Multiple Choice Given that the intercepts of a graph are ( − 4 , 0 ) and ( 0 , 5 ) , choose the statement that is true The y -intercept is − 4 , and the x -intercept is 5 . The y -intercepts are − 4 and 5 . The x -intercepts are − 4 and 5 . The x -intercept is − 4 , and the y -intercept is 5 .
Solution Summary: The author explains the intercepts of a graph that are (-4,0) and
The OU process studied in the previous problem is a common model for interest rates.
Another common model is the CIR model, which solves the SDE:
dX₁ = (a = X₁) dt + σ √X+dWt,
-
under the condition Xoxo. We cannot solve this SDE explicitly.
=
(a) Use the Brownian trajectory simulated in part (a) of Problem 1, and the Euler
scheme to simulate a trajectory of the CIR process. On a graph, represent both the
trajectory of the OU process and the trajectory of the CIR process for the same
Brownian path.
(b) Repeat the simulation of the CIR process above M times (M large), for a large
value of T, and use the result to estimate the long-term expectation and variance
of the CIR process. How do they compare to the ones of the OU process?
Numerical application: T = 10, N = 500, a = 0.04, x0 = 0.05, σ = 0.01, M = 1000.
1
(c) If you use larger values than above for the parameters, such as the ones in Problem
1, you may encounter errors when implementing the Euler scheme for CIR. Explain
why.
#8 (a) Find the equation of the tangent line to y = √x+3 at x=6
(b) Find the differential dy at y = √x +3 and evaluate it for x=6 and dx = 0.3
Q.2 Q.4 Determine ffx dA where R is upper half of the circle shown below.
x²+y2=1
(1,0)
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Discrete Distributions: Binomial, Poisson and Hypergeometric | Statistics for Data Science; Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=lHhyy4JMigg;License: Standard Youtube License