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Tuesday, May 22, 2012

007 James Bond - SKYFALL - Official Teaser Trailer



Tuesday, April 3, 2012

Indian Stock Market Story

Deepak Shenoy has a great chart which he uses to conclude that all of Indian Stock Market returns can be explained by 4 years 2003-2007. Here is the chart.

(Source: CapitalMind.in)

However, the chart only tells a part of the story. Here is the same chart along with earnings growth. I have normalized both earnings and price to 100 in the beginning so that you can clearly see the relationship.

(Data Source: BSEIndia.com)


As you can see that prices have tracked earnings over the last 20 years (1991 to 2012). So over the long term your returns in the stock market will depend on earnings growth, which in turn will depend on how the economy is doing. India’s GDP averaged over 9% from 2003 to 2007 and around 5.3% from 1991 to 2002. Attached is a chart of India’s GDP growth during this period.

(Source: Google Public Data & World Bank)


Earnings Growth (left axis) and GDP Growth (right axis) is attached here.


Let me recap, over long terms prices track earnings which tracks GDP growth. Thus, expected returns will more or less depend on GDP growth.

However it is important to note this is expected long term returns and should not be confused with you actual point to point return. In addition to earnings growth point to point return will also depend on PE multiple expansion (or compression). I had done a detailed post on this earlier.

This brings me back to Deepak’s chart. From May 2003 to Dec 2007 the market went from 3,180.25 to 20,286.99 (up 7 times) and PE went from 13.21 to 26.94. If the PE multiple remained the same (i.e. had markets tracked earnings) then the market should be at 9,947.70 (up 3 times). Thus, PE expansion was responsible for most of the returns we saw.

Similarly, from Dec 2007 to Apr 2012 the market went from 20,286.99 to 17,597.42 and PE multiple went from 26.94 to 17.92. If the PE multiple had remained the same the market would have been at 26,455.05. Here is a chart of Price, Earnings and PE. Again Price and Earnings have been normalized at 100 in 1991.



Thus, point to point returns will depend on valuations in addition to expected returns. If the stock market goes from undervalued to overvalued (as it happened from 2003 to 2007) your returns will be higher than expected returns and vice versa.

Currently market PE is around 17 whereas historically it has been around 21 and median is around 19 (averages are impacted by high PE multiples resulting from early 90s scams and 2000s bubble so median may be a better approximation). This indicates that markets are slightly undervalued. So (and I mean over 5 to 10 years) when PE expands to its historical value (around 19) then returns will be higher than what you expect.

This post can be reduced to following mathematical equations:
Point to Point Return = Expected Future Returns + Returns due to Multiple Expansion
Expected Future Return = Earnings Growth ~ ~ GDP Growth

Finally, though I haven’t discussed it in this post but it can be shown that
GDP Growth = Population Growth + Productivity Increases

Valuations will tell you what will happen to PE multiple (e.g. I would expect it to go up from here).
Population Growth is really predictable and is around 1.3% for India.
So, to predict returns all you really need is Productivity Increase Smile.
Let me know if you can predict it Open-mouthed smile.

Sunday, April 1, 2012

Hussman Interview Q3

Please see solutions to Q1Q2.

3) A vehicle is filled with fuel. One gallon of fuel can propel every 10 pounds of vehicle weight a
distance of 2 miles (5 pounds of weight 4 miles, etc). A gallon of fuel weighs 8 pounds.


a) Write a general expression for the distance traveled by the vehicle

b) If f0 is the starting amount of fuel, write an analytic equation for the total distance traveled

c) If the vehicle weighs 100 pounds when empty, and has traveled 1 mile after expending half
of its starting amount of fuel, what is the starting amount of fuel, and how far does it
ultimately take the vehicle (round to 4 decimal places)?


d) Suppose you didn’t discover that an analytic solution to this problem requires calculus (now
there’s a hint!). Write a short program in C++, Basic, Matlab or pseudocode to approximate
the amount of fuel required to move the vehicle (100 pounds when empty) 5 miles.


Weight of Vehicle = V; Weight of Fuel/Gallon = F; Gallons of Fuel = G; Distance Travelled = D; Weight of Vehicle and Fuel = W

I am not sure the difference between part a and b but here is my attempt.

a) D is directly proportional to G and inversely proportional to W
D ~ G/W => D = KG/W, where K is proportionality constant
In this case when D = 2; G = 1; W = 10. Therefore, K = 20.
Thus, when W = 5; D = 20*1/5 = 4

b) W = V + F*G => D = KG/(V+FG)
Total distance travelled is tricky because as soon as you burn a little fuel the weight reduces and the further you can travel.
Therefore, to find distance travelled we first need to know what happens when fuel changes by small amount i.e. differentiate the above equation with respect to G
clip_image002[10]
Now to find total distance travelled we have to integrate when fuel G ranges from f0 to 0
Let us first solve indefinite integral
clip_image002[20]
Now adding limits total distance travelled is
clip_image002[22]
 
c) Weight of Vehicle = V = 100lbs; Weight of Fuel/Gallon = F = 8lbs/g; Gallons of Fuel = G = f0 –> f0/2; Distance Travelled = D = 1 and K = 20
clip_image002[24]
Substituting,
clip_image002[26]
10000+1200f0+32f02 = 250*(-4f0)
32 f02 + 2200f0 + 100 = 0
f0 ~ –63.8562 & –4.8938
The equation has 2 roots and both negative. So, it is clear I have made a mistake. However, my calculus is a little rusty and I will leave the solution so that the internet can explain the mistake Smile
 
d) This will only make sense if I have the correct equation. So, I will not attempt to solve it.

Hussman Advisors Financial Analyst Pre Interview Question 2

I am solving question 2 of Hussman Advisors Financial Analyst pre interview question. Question 1 was solved earlier.

Q2 Let the value of a firm’s equity be equal to the discounted stream of deliverable cash flows
(“DCF”) and assume that firm
wide DCF grows at rate g, so Vt = DCFt+1 / (kg). Let Nt be the
number of shares of stock at time t, so the per
share stock price Pt is simply Vt/Nt. Each year, the
company pays a proportion d of DCF to existing shareholders as dividends, and then uses the
remaining proportion (1
d) to repurchase shares. Assume that shares are repurchased at price
Pt, which simultaneously determines Nt.



a) Derive a simple expression for the pershare dividend growth rate g*, in terms of k, d and g
only, and give a simple interpretation (Hint: Write down expressions involving DCF, don’t be
afraid to substitute equivalent expressions, and notice that all growth rates will be constant
given the assumptions above. In particular, Nt/Nt+1 = Nt
1/Nt).


b) Show algebraically that if the pershare dividend growth rate g* is used, the standard
dividend discount model holds even in the presence of share repurchases, so that adding in
repurchases as if they were a separate payment to shareholders would actually represent
double
counting. Specifically, show that Vt/Nt = Dt+1 / kg*.

a) From the problem statement we have the following equation:

P0 = V0 / N0 = (DCF1/N0) / (k-g)

I will use g’ instead of g* as * is generally used as multiplication sign and may confusion i.e. g’ = g*
We can also drive the price per share using dividend discount method i.e. if Dividends Dt grow at constant rate g’ then

P0 = D1/k-g’

Also, from the problem statement Dt = d*DCFt
Therefore, P0 = D1/k-g’ = (d*DCF1/N0) / (k-g’)

The price per share obtained from both the methods should be same.

Therefore, k-g’ = d*(k-g) or g’ = g* = k - d*(k-g)

I am not sure if the question expected me to prove that P0 obtained from two methods is same or not.
However, there is an entire paper that goes on to show the equivalence of Cash Flow Method and Dividend Discount Method.


b) From the result above: d * (k-g) = (k-g’)

Vt/Nt = DCFt+1/(k-g) = Dt+1 / d*(k-g) = Dt+1 / (k-g’)

This makes sense since we assumed Pt (Vt/Nt) is equivalent to derive g’ in the first case.

Hussman Strategic Advisors Financial Analyst Pre Interview Questions

Hussman Strategic Advisors has a bunch of pre-interview questions for Financial Analyst / Financial Engineer position. I will try my best to answer some of those questions. I will tackle the 1st question in this post.

What is the payment on a 12‐year, $400,000 mortgage that compounds on the payment date every 73 days at 6%? (show your formula and calculations, and ignore leap year)

I do not really understand the question so here is my best guess. I am assuming I do not have to worry about individual days in a month. Also, I am assuming 6% is the annual rate.


Step 1: Calculate relationship between annual rate, 73 days rate and monthly rate.

365 Days = 12 months => 73 Days = 12*(73/365) months = 2.4 months

Let R be the annual rate in decimal (0.06 in this case); r be the monthly rate in decimal and R* is the rate every 73 days (or every 2.4 months)

To convert rates lets start with something familiar.
If a loan compounds every 6 months (half yearly) then 6 months rate is R/2 i.e. R/(12/6)
If a loan compounds every 3 months (quarterly) the 3 months rate is R/4 i.e. R/(12/3)
If a loan compounds every month then monthly rate is R / 12 i.e. R/(12/1)

So, if a loan compounds ever 2.4 months (73 days) the 73 day rate is R/(12/2.4) = R/5 = R*

Now, monthly rate r compounded for 2.4 months = R* i.e. (1+r)2.4 = 1 + R*

r = (1+R*)(1/2.4) – 1 = (1+(R/5))(1/2.4) – 1 = (1+(0.06/5))(1/2.4) – 1 = 0.004983 = 0.4983%

If the loan had compounded every month instead of 73 days then r = R/12 = 0.5%


Step 2: Derive Payment Formula

Let L be the loan amount ($400,000) and P be the payment every month

At the end of the 1st month:
Outstanding Loan Amount, L1 = L*(1+r) – P
Let (1+r) = X
L1 = L*X – P

At the end of 2nd month:
Outstanding Loan Amount, L2 = L1*X – P = L*X2 – P*X – P

At the end of 3rd month:
Outstanding Loan Amount, L3 = L2*X – P = L*X3 – P*(X2 – X – 1)

At the end of nth month:
Outstanding Loan Amount, Ln = Ln-1*X – P = L*Xn – P*(Xn-1 – Xn-2 – … 1)

Sum of Geometric Progression (GP) 1+x+x2+x3+…xn-1 = (xn – 1)/(x – 1)
Using the GP formula and replacing X by (1+r)

Ln = L*(1+r)n – P*(((1+r)n – 1) / r)

At the end of loan term (n = 144 months in this case) outstanding amount = 0
When Ln = 0 then P = L*r*(1+r)n / ((1+r)n – 1) = L*r / (1 – 1/(1+r)n)
Therefore, P = 400,000*0.004983 / (1-1/(1+0.004983)144) = 3899.39