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cpcoolestkid4
12.03.2021 •
Business
Consider the following data for a one-factor economy. All portfolios are well diversified.
Portfolio E(r) Beta
A 12% 1.2
F 6% 0.0
Suppose that another portfolio, portfolio E, is well diversified with a beta of .6 and expected return of 8%. Would an arbitrage opportunity exist? If so, what is the arbitrage strategy?
When beta = 0, there is no risk, so it is risk free
Since beta = 0, the expected return for Portfolio F equals the risk-free rate
For Portfolio A, the ratio of risk premium is (12% – 6%)/1.2
.12 - .06/1.2 => .06/1.2 = .05 x 100 = 5%
For Portfolio E, the ratio is lower at (8% – 6%)/.6
.08 - .06/.6 => .02/.6 = .0333 x 100 = 3.33%
This implies that an arbitrage opportunity exists
Please explain to me why an arbitrage opportunity exists
Find out weight for Portfolio G:
W1: Weight in Portfolio A
W2: Weight in Portfolio G
W2 = 1 – W1
W1 B1 + (1 – W1)B2 = .6
W1 1.2 + (1 – W1)0 = .6
1.2W1 + 0 = .6
1.2W1 = .6
W1 = .6/1.2 = .5
Expected Return and Beta of Portfolio G:
E(rG) = (.5 x 12%) + (.5 x 6%)
E(rG) = (.5 x .12) + (.5 x .06)
E(rG) = .06 + .03 = .09 x 100 = 9%
βG = (.5 x 1.2) + (.5 x 0)
βG = .6 + 0 = .6
Comparing Portfolio G to Portfolio E, G has the same beta, but a higher expected return than E. Therefore, an arbitrage opportunity exists by buying Portfolio G and selling an equal amount of Portfolio E. The profit for this arbitrage will be:
rG – rE = [9% + (.6 x F)] – [8% + (.6 x F)] = 1%
1% of the funds (long or short) in each portfolio
I believe the answer is correct, but I would like to know why an arbitrage opportunity exists. Explain that to me in relation to the problem and please solve this step by step to get 1% rG – rE = [9% + (.6 x F)] – [8% + (.6 x F)] = 1%
Solved
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Ответ:
The atoms of a substance can be determined by dividing the given mass of the substance by the molar mass. This results to moles of the substance. Then we multiply by the Avogadro's number. In this case, we divide 1 g by 58. 69 g/ mol and multiply by 6.022 x10^23 atoms / mole. The answer is 1.03 x10^22 atoms
Explanation: