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Thread: A Model of Investment Gearing

  1. #1
    Newbie emterics90's Avatar
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    A Model of Investment Gearing

    I have made up by myself a mathematical model of gearing that is supposed to help me decide whether I should gear or not. I am still young and not a finance PhD, so I would like someone to see whether my model is correct or whether I have made some serious errors.

    Gearing means borrowing to invest. If I borrow to invest, then my profit from gearing $\displaystyle \pi_G$ is

    $\displaystyle \pi_G = L(1+r)^T - R \int^{T}_{0} (1+r)^x dx$

    where L is the amount you loan from the bank, r is the annual rate of growth of your investment, T is the duration of your loan, and R is the yearly interest repayment.

    The first term $\displaystyle L(1+r)^T$ is the value of your investment after T years and the second term $\displaystyle R \int^{T}_{0} (1+r)^x dx$ is the opportunity cost of your investment because you could have invested without borrowing.

    Solving the integral, we get,

    $\displaystyle \pi_G = L(1+r)^T - \frac{R(1+r)^T}{log(1+r)}$

    To see whether you should gear, find out whether $\displaystyle \pi_G > 0$. If it is, borrow to invest. Otherwise, don't.

    By looking at the formula we can see that gearing is a better option if L (the loan amount) is high and R (the interest repayments) is low.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by emterics90 View Post
    I have made up by myself a mathematical model of gearing that is supposed to help me decide whether I should gear or not. I am still young and not a finance PhD, so I would like someone to see whether my model is correct or whether I have made some serious errors.

    Gearing means borrowing to invest. If I borrow to invest, then my profit from gearing $\displaystyle \pi_G$ is

    $\displaystyle \pi_G = L(1+r)^T - R \int^{T}_{0} (1+r)^x dx$

    where L is the amount you loan from the bank, r is the annual rate of growth of your investment, T is the duration of your loan, and R is the yearly interest repayment.

    The first term $\displaystyle L(1+r)^T$ is the value of your investment after T years and the second term $\displaystyle R \int^{T}_{0} (1+r)^x dx$ is the opportunity cost of your investment because you could have invested without borrowing.

    Solving the integral, we get,

    $\displaystyle \pi_G = L(1+r)^T - \frac{R(1+r)^T}{log(1+r)}$

    To see whether you should gear, find out whether $\displaystyle \pi_G > 0$. If it is, borrow to invest. Otherwise, don't.

    By looking at the formula we can see that gearing is a better option if L (the loan amount) is high and R (the interest repayments) is low.
    Would it not make more sense to redure R to an annual eqivalent rate R',
    then this is worth doing if R' < r.

    RonL
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  3. #3
    Newbie emterics90's Avatar
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    R is the yearly interest repayments. So for example suppose you buy a house and the bank lend you $100,000 so then L=100,000. The bank requires you to repay $500 a month, and so $\displaystyle R = 500 \times 12$.

    I'm a little uncertain about the second term, the integral.

    Maybe it looks like this. If I buy a house and get L from the bank and have to pay back R per year then after T years I will have (assuming the houses appreciates by r per year)

    $\displaystyle
    \pi_G = L(1+r)^T - RT
    $

    where $\displaystyle L(1+r)^T$ is the price of the house after T years and RT is cost of the interest repayments. $\displaystyle \pi_{G}$ is the profit from gearing. $\displaystyle \pi_{NG}$ is the profit from not gearing.

    If however instead of buying a house I use the money I would otherwise use for interest repayments for a house to buy, say, shares that go up at rate r per year then I get the following:

    $\displaystyle
    \pi_{NG} = R(1+r) + R(1+r)^2 + R(1+r)^3 + ... + R(1+r)^T
    = R \int^{T}_{0} (1+r)^x dx
    $

    And so I should get a loan from the bank to buy property or shares if $\displaystyle \pi_G - \pi_{NG} > 0$ or

    $\displaystyle
    L(1+r)^T - RT - R \int^{T}_{0} (1+r)^x dx > 0$

    Further mathematical manipulation of the above gets the following rule:

    Borrow to invest if

    $\displaystyle \frac{L}{R} > \frac{1}{\log(1+r)} + \frac{T}{(1+r)^{T}}$

    So for example suppose I wanted to be as wealthy as possible in 11 years time, so I make T=11. Returns on property or shares in the long run is about 9% so I will set r=0.09. Plugging these into the equation gives the following:

    Borrow to invest if

    $\displaystyle \frac{L}{R} > 15.866$

    Remember L is the loan amount and R is the yearly interest repayment. So suppose the bank offers to loan you $300,000 and asks for you to pay back $500 monthly (or $6000 yearly) then L=300,000 and R=6000 and $\displaystyle \frac{L}{R}=50>15.866$, so you should accept this deal from the bank and borrow to invest.

    Most people who buy houses tend to just buy it for emotional reasons, e.g. it looks nice, but I am trying to reduce here a rigid rule for whether or not taking out a mortgage is a good idea. Taking out a loan from the bank to invest needn't be for property investment. Investment in shares is also possible.
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