Category: Uncategorized

If I want to make an audio or a video recording during class,

If I missed test 1, test 2 and my final exam score is 85,

If I missed test 1 and my final exam score is 85,

Consider the function h = lambda x: x**2 + 4*x – 5 / x If we use fsolve from scipy.optimize to find the x where h(x) = 0, with a starting guess of x0=0.01, what is the output we obtain?

Consider the pseudo code below to obtain the efficient portfolios:from scipy.optimize import minimize f = lambda w: TO BE FILLED mu = np.linspace(15, 30, 31) sd_optimal = np.zeros_like(mu) w_optimal = np.zeros([31, 5]) for i in range(len(mu)): # Optimization Constraints cons = ({‘type’:’eq’, ‘fun’: lambda w: np.sum(w) – 1}, {‘type’:’eq’, ‘fun’: lambda w: w @ ER * 252 * 100 – mu[i]}) result = minimize(f, np.zeros(5), constraints=cons) w_optimal[i, :] = result.x sd_optimal[i] = np.sqrt(result.fun)Assuming that ER are Cov given, what should we substitute TO BE FILLED for in order to get the desired result?

What is the output of the following code? newlist = np.zeros(4) for i in range(len(newlist)):    newlist[i] = i + newlist[i] print(newlist)   

Consider a portfolio with the following weights w,  expected return on each risky asset ER, and covariance matrix Cov below. w = np.array([0.05, 0.03]) ER = np.array([0.10, 0.02])Cov = np.cov([[0.004, 0.0156], [0.0156, 0.009]]) Which of the following expressions represents the portfolio volatility in Python?

How many atoms are in a face centered cubic lattice?

Consider the code below with numbered lines:1)def h(x): 2) return np.exp(-x**2 / 2) / np.sqrt(2 * np.pi) 3) 4)x = np.linspace(-4, 4, 51) 5)y = np.zeros(x) 6) 7)for i in range(len(y)-1): 8) y[i] = h(x[i]) 9)plt.plot(x, y) If we run the code above, we will receive an error. In which line lies the error?

Assume we live in a world where the CAPM holds and it is given by  ER = lambda beta: beta * 0.05 + 0.02If you know this linear relationship between beta and ER, you can compute the ER for a given beta.What is the ER on a stock with beta = 0.5?