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nydimaria [60]
3 years ago
10

f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected

performance, what fraction of your portfolio should be allocated to T-bills and what fraction to equity? Assume your utility function is U = E(r) – 0.5 × Aσ2. and 51.05%

Physics
1 answer:
siniylev [52]3 years ago
7 0

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

Weight=\frac{E_M-r_f}{A\sigma_M^2}

For the equity, here as

  • Weight is percentage of the equity which is to be calculated
  • {E_M-r_f} is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
  • A is the risk aversion factor which is given as 4.
  • \sigma_M is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%

So the weight of equity is 48.94%.

Now the weight of T bills is given as

Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\

So  the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

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3 years ago
A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w
sveta [45]

Answer:

A)     t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

           v = v₀ - g t

the highest point zero velocities (v = 0)

           t = (v₀-v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

to calculate the height let's use

          v² = v₀² - 2 g y

          y = vo2 / 2g

           y = 4 2 / (2 9.8)

          y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

          # _photos1 = 0.40916 (30/1)

          # _photos1 = 12

yes i take 120 fps

          #_fotod = 0.40916 (120/1)

          #photos = 5.87 10³

 

B) The ball is released from a latura h how long it takes to reach the floor

           v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

            v² = 2gy

             v = √ (2 9.8 0.916)

             v = √ (2.1397 101)

             v = 4.6257 m / s

c) we ask us for the time for latura

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

now we can use the formula

             y = v₀ t - ½ g t²

           0.458 = 4.00 t - ½ 9.8 t²

            4.9 t² - 4t + 0.458 = 0

            t² -0.8163 t +0.09346 = 0

we solve second degree execution

           t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

           t = [0.8163 ± 0.540] / 2

           t₁ = 0.678 m

           t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

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