Archive for the ‘Maths’ Category

Nominal and real interest rates

July 28, 2011 Leave a comment

One of the most simple but crucial aspects of learning economics is understanding the relationship between nominal and real interest rates. When you see interest rates quoted for savings accounts or loans, they are usually nominal interest rates, expressed in terms of currency. Eg if you borrow £1000 for one year at 4%, then you pay back £1040, you are paying in interest 4% of the nominal value of the amount you borrowed.

However that does not tell you how the £1040 you repay compares to the £1000 you borrowed in terms of a basket of goods. If inflation was 3.5% then £1040 in a year’s time will be worth \frac{1040}{1.035}=1004.83 ie £1004.83 today. So you are actually only really paying \frac{1004.83}{1000}=1.00483 ie 0.483% interest in real terms.

This is a concept that most people in the real world won’t think about. They will just think about the nominal rate of interest quoted and not take account of how this will be eroded by inflation.

The relationship between real interest rates, nominal interest rates and inflation is

(1+r_t) = \frac{(1 + i_t)}{1 + \pi^e_{t+1}}.

This means that the real interest rate depends on the nominal interest rate and the expected rate of inflation next year. The real interest rate is an ex ante interest rate, because it is based on expectations of inflation. This means at the time you are making a decision (do I save/borrow at this nominal rate of interest?) you have to base your decision on what your expectations of inflation are. A year later, when you know what inflation actually was, you can find out the ex post real interest rate (ie what the real interest rate actually was, regardless of what the ex ante real interest rate suggested it would be).

You can make an approximation to this.

(1+r_t) = \frac{(1 + i_t)}{1 + \pi^e_{t+1}} \Rightarrow (1 + r_t) (1 + \pi^e_{t+1}) = (1+i_t), when you expand the brackets you get:
1+r_t + \pi^e_{t+1} + r_t \pi^e_{t+1}  = 1 + i_t \Rightarrow r_t + \pi^e_{t+1} + r_t \pi^e_{t+1}  = i_t .

As the multiple of real interest rate and expected inflation r_t \pi^e_{t+1} is likely to be small, we can approximate to:

 r_t + \pi^e_{t+1} \approx i_t or  r_t \approx i_t - \pi^e_{t+1}.

Real interest rates are relevant when we are thinking about what something is worth in terms of a basket of goods, so for instance firms that are making decisions about whether or not to make an investment, will think in terms of real interest rates. The real interest rate is the relative price of current consumption compared to future consumption. It tells you how much consumption you gain in the future by sacrificing consumption now.

The IS relation uses the real interest rate: Y=C(Y-T) + I(Y,r) + G + NX. As  r \approx i - \pi^e we can roughly say Y=C(Y-T) + I(Y,i - \pi^e) + G + NX. This is how expected inflation influences the IS curve. A rise in expected inflation will mean that a higher nominal interest rate is needed for every level of output. So a rise in expected inflation will shift the IS curve upwards.

Nominal interest rates are relevant when we are thinking about money markets. When investors are deciding whether to hold money (that pays 0% interest) or some illiquid interest bearing asset like bonds (that pays i% interest) then they will think in terms of nominal interest rates.

The LM relation uses the nominal interest rate: \frac{M}{P}=YL(i).

Categories: Simple algebra

Deriving the demand elasticity for a competitive firm

June 21, 2011 Leave a comment

This is a neat little exercise in using differentiation in economics and getting used to the notation involved, and using tricks of algebraic manipulation such as multiplying top and bottom of a term in an equation by the same value.

The goal here is to find an expression for the elasticity of demand for a firm in a competitive market, in terms of the market elasticity of demand and supply.

The basic concept here is one of residual demand curves. When there are a number of firms in the market, you will have a market demand curve which tells you how much of the good consumers will produce at a given price, and each firm will face a residual demand curve, which is the demand that is left over for them that is not met by other sellers. We are going to assume here that all the firms are identical, face identical marginal costs and produce the same output.

You can write it like this D^r(p) = D(p) - S^o (p) (1)

This notation basically says the residual demand function equals the market demand function minus the supply function of all the other firms. These demand and supply firms are really in terms of output, but if you used the notation Q(p) for both you would get mixed up later on in the algebra. The reason p is in brackets is that all of these functions are written in terms of p.

Small letters, p and q, get used for a firm’s price and quantity, while capital letters P and Q will refer to market price and quantity. Of course as we are talking about a competitive market here, p and P will be the same, as all firms are ‘price takers’, they face the same price. q=\frac{Q}{n} if there are n identical firms all producing the same amount of output. The output produced by others is Q_o=(n-1)q which is basically saying there are (n-1) ‘other’ firms (aside from the firm we are looking at), each producing q output.

If we want to express this in terms of elasticities, it’s useful to think of some definitions before we start.

The market elasticity of demand is \epsilon = \frac{p}{Q} \frac{dQ}{dp}. Now in equation (1) above the D(p) is really Q(p) as the market demand function is in terms of Q, the notation D and S just gets used so you don’t get mixed up with which Q is for what later one. So with respect to our equation, \epsilon = \frac{p}{Q} \frac{dD}{dp}.

The elasticity of demand which a firm faces is \epsilon_{firm} = \frac{p}{q}\frac{dD^r}{dp}. This time we are using q instead of Q because it refers to the firm’s output not the market output, and we are using the differential of the residual demand curve rather than the market demand curve.

The elasticity of supply from other firms is \eta_{o} = \frac{p}{Q_o}\frac{dS^o}{dp}. This time you are thinking in terms of the quantity supplied by other firms, and the differential of the supply function of other firms.

Now given these definitions we can work towards them starting by differentiating equation (1) with respect to p.

\frac{dD^r}{dp} = \frac{dD}{dp} - \frac{dS^o}{dp}
Again this is the type of differentiation notation you will get used to in economics. You aren’t actually given a function to differentiate here, you are just expressing what the differential is!

Now for some tricks to manipulate this equation. First multiply throughout by \frac{p}{q}

\frac{p}{q}\frac{dD^r}{dp} = \frac{p}{q}\frac{dD}{dp} - \frac{p}{q}\frac{dS^o}{dp}
This has allowed us to write the left hand side in terms of the firm’s demand elasticity, from the definition above

\epsilon_{firm} = \frac{p}{q}\frac{dD}{dp} - \frac{p}{q}\frac{dS^o}{dp}

Now you might be able to see where we are going with this. That first term on the right hand side is nearly the market elasticity of demand, except on the denominator we have q instead of Q. So multiply top and bottom of that term, by Q (which is just multiplying by 1 as Q divided by Q is 1, so it doesn’t change it).

\epsilon_{firm} = \frac{p}{Q}\frac{dD}{dp}\frac{Q}{q} - \frac{p}{q}\frac{dS^o}{dp}

So now it is in terms of the market elasticity of demand.

\epsilon_{firm} = \epsilon \frac{Q}{q} - \frac{p}{q}\frac{dS^o}{dp}.

That second term on the right hand side is nearly the elasticity of supply from other firms, but again we have the wrong denominator, we need to get Q_o in there, so lets multiply top and bottom by Q_o

\epsilon_{firm} = \epsilon \frac{Q}{q} - \frac{p}{Q_o}\frac{dS^o}{dp}\frac{Q_o}{q}.

Now we have the elasticity of supply from other firms in there.

\epsilon_{firm} = \epsilon \frac{Q}{q} - \eta_{o}\frac{Q_o}{q}.

Remember above we said that q=\frac{Q}{n} and Q_o=(n-1)q. This means our equation becomes

\epsilon_{firm} = \epsilon \frac{Q}{\frac{Q}{n}} - \eta_{o}\frac{(n-1)q}{q}

So this cancels down to

\epsilon_{firm} = n\epsilon - \eta_{o}(n-1)

That’s how you express the elasticity of demand for a competitive firm in terms of the market elasticity of demand, the elasticity of supply of other firms, and the number of firms in the market.

Optimisation with a constraint

June 20, 2011 Leave a comment

In A level maths you will be used to solving standard optimisation problems like y = -3x^2 + 18x - 4 find the value of x that gives you the maximum value of y. You just differentiate it and set it equal to zero, ie \frac{dy}{dx} = -6x + 18 so when 0 = -6x + 18, x = 3 so y = 23.

In economics you do a lot of optimisation but a lot of the time you are working within a budget constraint.

For example suppose a firm has a production function, Q = 2K^{0.3}L^{0.7}

This shows you how the output is related to the inputs of capital (K) and labour (L), so if you have 10 units of capital and 12 units of labour you will get Q = 2(10^{0.3})(12^{0.7})=22.723.

Now if we wanted to ‘optimise’ our output, and we had an unlimited budget, then we could just keep buying more capital and hiring more labour and our output would go up and up and we could reach any level of output we wanted. However in the real world, there will be a cost for capital and a cost for labour.

Suppose a unit of capital costs 2 and a unit of labour costs 3, and we have a total budget of 30. There are lots of ways we could use that budget, we could afford 3 units of capital and 8 units of labour and have an output of 11.921, or we could afford 6 of each and have output of 12 etc. How do we know what is the ‘optimum’ way to spend our budget to maximise output given that constraint? This is a very ‘real world’ problem that you can use mathematics to solve.

You solve this using something the Lagrange multiplier which is something you probably won’t have been taught at A-level, but all you need to know is partial differentiation and you can do it.

I won’t go into the full mathematical reasoning behind why it works here, just show you how to do it.

Basically you have to optimise an original function: f(x,y) subject to a constraint:\phi(x,y)=M

So you set up a new function: g(x,y,\lambda)=f(x,y)+\lambda(M-\phi(x,y)) and find the three partial derivatives and set them equal to zero, ie \frac{\partial g}{\partial x} = 0, \frac{\partial g}{\partial y} = 0, \frac{\partial g}{\partial \lambda} = 0. Then you solve the simultaneous equations and you have your values for x and y.

It will make more sense if we look at it in terms of this example above:

Objective function is: Q = 2K^{0.3}L^{0.7}
Constraint is 2K + 3L = 30

So we set up our new function: g(K,L,\lambda)=Q = 2K^{0.3}L^{0.7}+\lambda(30-(2K + 3L))

Now we do our three partial derivatives and set them equal to 0:

\frac{\partial g}{\partial K} = 0.6K^{-0.7}L^{0.7}-2\lambda = 0 (1)
\frac{\partial g}{\partial L} = 1.4K^{0.3}L^{-0.3}-3\lambda = 0 (2)
\frac{\partial g}{\partial \lambda} =30 - 2K - 3L = 0 (3)

Now we can solve our simultaneous equations…
First put equation (1) and equation (2) in terms of \lambda

So  0.3K^{-0.7}L^{0.7}=\frac{1.4}{3}K^{0.3}L^{-0.3}

We can multiply both sides of this equation by K^{0.7} and we get  0.3L^{0.7}=\frac{1.4}{3}KL^{-0.3}

Now we can multiply both sides of this equation by L^{0.3} and we get  0.3L=\frac{1.4}{3}K so  L=\frac{1.4}{0.9}K

Now we can substitute this into equation (3) to get it all in terms of K, so 30 - 2K - 3(\frac{1.4}{0.9}K) = 0
So 30 - 2K - 3(\frac{1.4}{0.9}K) = 0 \Rightarrow 30 = 2K + 3(\frac{1.4}{0.9}K)\Rightarrow 30 = \frac{20}{3}K so K = 4.5

We can now put this back into our expression for L to get  L=\frac{1.4}{0.9}4.5 so L = 7

So our optimum choice is to spend our budget of 30 on 4.5 units of capital and 7 units of labour, this will give us a total output Q of 12.262. No other combination, within our budget constraint, would reach a higher value of output.

A nice practical application of partial differentiation to solving an economic problem.

Categories: Maths, Optimisation


June 20, 2011 Leave a comment

This is something which you will have come across in A-level, basically optimising a function by setting the derivative (or partial derivative) to 0.

In economics you are optimising things all the time. Firms are looking to optimise revenues or profits, or minimising costs. You will be given demand functions and information on costs etc, and asked to optimise.

For instance, say you have a monopoly firm facing a demand function Q = 200 - 2P, you have fixed costs of 25 and produce at a marginal cost of 5 per unit. You are asked to optimise profits.

From the information here we need to set out our profit function: \pi = TR - TC

First write the demand function in terms of P, the ‘inverse demand’ function: P = 100 - 0.5Q. This helps when we want to get our total revenue function. TR = PQ so here TR = 100Q - 0.5Q^2.

Now lets think of the costs. TC = FC + MC(Q) and we have fixed costs of 25 and marginal cost of 5, so here TC = 25 + 5Q

So now we have an expression for profit, \pi = TR - TC here means \pi = 100Q - 0.5Q^2 - 25 - 5Q so \pi = 95Q - 0.5Q^2 - 25. We optimise profit by setting \frac{d\pi}{dQ}=0, here \frac{d\pi}{dQ}=95-Q so setting it equal to 0 means our profit maximising output will be 95.

You could also have solved that by finding marginal revenue and setting it equal to marginal cost because the monopolist will produce where MR = MC.

We get marginal revenue by differentiating total revenue with respect to output, MR = \frac{d(TR)}{dQ}, here MR = 100 - Q. So if MC is 5 we know that at the production point MR is 5. Putting that into the equation for MR means we have Q of 95.

So our firm is producing an output of 95, which means when you sub that into the original demand function you get a price of 52.5. We can work out the actual value of the profits by going back to our expression for profit, \pi = 95Q - 0.5Q^2 - 25, which gives profits of 4487.5 at the profit-maximising output.

What if you were asked to find the revenue maximising point rather than profit maximising point? You had TR = 100Q - 0.5Q^2 so this time you just set \frac{d(TR)}{dQ}=0. Here \frac{d(TR)}{dQ}=100 - Q so when you set it equal to 0 you get a revenue maximising point of 100. This is of course where the marginal revenue is 0.

If you increase output beyond that point you are getting a negative marginal revenue. Why would this be? Well any firm that faces a downward sloping demand curve will have to decrease its price, if it wants people to buy more of its good. So it has a trade off – it sells more output, but each unit is being sold at a lower price. There comes a point where you lose more by reducing the price on all the units you are selling, than you gain by selling an extra unit, and that’s when you’ve gone past the point where MR = 0. Here, if you can sell 100 units at a price of 50, then you bring in revenue of 5000. But in order to sell 101 you have to reduce the price (this is just reading off the original demand function) to 49.5. This means you get revenue of 4999.5, so your revenue has dropped because you are selling beyond the revenue maximising point.

Categories: Maths, Optimisation


June 20, 2011 Leave a comment

This is something which you will see a lot. The elasticity gives you a measure of proportionate response from one variable to another, it is commonly used for price and quantity. The elasticity of demand tells you how much quantity demanded changes when you change the price. Inelastic demand means you can raise the price and not suffer too much of a drop in output. Elastic demand means if you raise the price, people stop buying it and you suffer a larger proportionate drop in output. Obviously firms prefer to have inelastic demand. Various things affect elasticity of demand, the availability of substitutes is a big one.

Elasticity of demand is the proportional change in quantity demanded divided by the proportional change in price: \frac{\triangle Q}{Q}\div\frac{\triangle P}{P} = \frac{\triangle Q}{Q}\frac{P}{\triangle P} which is generally expressed as \frac{P}{Q}\frac{\triangle Q}{\triangle P}. This will generally be a negative value, because when you increase the price, you decrease the quantity, so \triangle Q will be negative.

As these changes tend to zero (ie at the margin) we can express the elasticity as \epsilon=\frac{P}{Q}\frac{dQ}{dP}.
If -1<\epsilon<0 then we say the demand is inelastic. If \epsilon=1 then it is 'unit elastic'. If -\infty<\epsilon<-1 then it is elastic.

Another little manoeuvre with elasticity involves using the chain rule again. Remember that a demand function expresses Q as a function of P. If we take the natural log of Q then we can differentiate it with respect to P by the chain rule:\frac{d}{dP} ln (Q(P)) = \frac{1}{Q}\frac{dQ}{dP}. You differentiate the outer function first (which is the natural log of Q) and multiply it by the differential of the inner function (which is Q in terms of P). At A-level you tend to get chain rule questions when you have to actually solve equations with values in them, but quite often in economics you get them written in this function notation so you have to get the hang of doing the chain rule that way.

Anyway since we just found that \frac{d}{dP} ln (Q(P)) = \frac{1}{Q}\frac{dQ}{dP} we can multiply both sides by P to get P\frac{d}{dP} ln (Q(P)) = \frac{P}{Q}\frac{dQ}{dP} which is the elasticity of demand again.

So \epsilon=P\frac{d}{dP} ln Q. This can be a useful way of calculating elasticities if the demand function is a bit complicated.

One area where this comes up is in the case of constant elasticity. Usually the elasticity will vary along different points of the demand curve (even if \frac{dQ}{dP} is constant, at every point there will be a different combination of P and Q so the \frac{P}{Q} part will vary. But there is a specific type of function that will give a constant elasticity all the way along the curve, it will look like this: Q=aP^{-b}

Now if we log both sides we will get ln Q=ln a - b lnP (this is using rules of logs).
So our formula above of \epsilon=P\frac{d}{dP} ln Q says that we need to differentiate that expression with respect to P, and multiply the whole thing by P. \frac{d}{dP}(ln a - b ln P)=\frac{-b}{P} so \epsilon=P\frac{-b}{P}=-b. So the elasticity will always be the constant -b. That's why constant elasticity functions are always of that form.

Differentiation in Economics

June 20, 2011 Leave a comment

Most undergrad level core micro and macro involves fairly simple differentiation, you will do a lot of optimisation and use the chain rule and product rules a lot. One thing you will have to get used to in economics is seeing things written as functions and differentiating them.

You are always differentiating to find ‘marginals‘. The concept of ‘marginals’ (marginal revenue, marginal product, marginal cost) etc is about the most important concept in microeconomics, because all decisions are taken ‘at the margin’. Do you increase production by another unit or just produce at the level you are doing? Well if your marginal revenue (the amount of revenue you will earn by producing another unit of output) is higher than your marginal cost (the amount it will cost you to produce another unit) then go for it. If your marginal cost is higher then you don’t. As you produce more your MR will fall and your MC will rise so you will maximise profits by producing where MR = MC. Basic golden rule of micro!

Because MR is basically the ‘change in revenue over the change in output’ you find it by differentiating total revenue with respect to output. Total revenue is price x quantity.

So you have TR = PQ, MR = \frac{d(TR)}{dQ} so MR = \frac{d(PQ)}{dQ}

Just a note here, don’t get confused by the fact that PQ is P times Q, and TR and MR are ‘total revenue’ and ‘marginal revenue’ you aren’t doing T times R or M times R, its just the abbreviation.

So if MR = \frac{d(PQ)}{dQ} you use the product rule to say MR = P\frac{dQ}{dQ}+Q\frac{dP}{dQ} so MR = P+Q\frac{dP}{dQ}

You also have to use the chain rule, and recognise how to do this in terms of the notation. Lets consider the question ‘what wage do you pay if you want to maximise profit?’

To show this example you first have to understand the concept of the production function. This will express Q as a function of K and L.

So we know that TR is a function of P and Q, and Q is a function of K and L. We have a ‘function of a function’ so the chain rule is coming up when we have to differentiate.

Now in terms of costs, we usually use w to refer to the cost (per unit) of labour (ie the wage) and r to refer to the cost of capital. If the capital was say machinery, then you aren’t having to pay the machinery a wage, but it is costing you, either you are renting it for a fee, or if you bought the machines outright, there is an ‘opportunity cost’ (ie you could have been investing the money you spent on the machines at the market interest rate, r, so if your machines are earning you less than r, you would have been better not buying them and investing your money instead)

So your total costs are going to be the wage times the number of units of labour you hire, plus the cost of capital times the number of units of capital you are using, ie TC = wL + rK

Your profits (usually \pi is used to symbolise profit in economics, are \pi = TR - TC so \pi = TR(Q(K,L)) - wL - rK. What this notation is saying is that total revenue is a function of Q (its also a function of P, but we don’t need that here as we are looking at labour which comes directly into the function for Q not P), and Q is a function of K and L.

To maximise profits with respect to the cost of our inputs, K and L, its an optimisation problem, we set the two partial derivatives of profit by K and L to zero to find the optimising points, ie \displaystyle \frac{\partial \pi}{\partial K} = 0 and \displaystyle \frac{\partial \pi}{\partial L} = 0.

We use the chain rule to do these partials. \displaystyle \frac{\partial \pi}{\partial K} = \frac{d(TR)}{dQ}\frac{dQ}{dK}-r and \displaystyle \frac{\partial \pi}{\partial L} = \frac{d(TR)}{dQ}\frac{dQ}{dL}-w. When you use the chain rule you do the derivative of the outside function times the derivative of the inside function.

Remember that \frac{dQ}{dK} = MR so \displaystyle \frac{\partial \pi}{\partial K} = MR\frac{dQ}{dK}-r and \displaystyle \frac{\partial \pi}{\partial L} = MR\frac{dQ}{dL}-w.

We can also introduce a couple of concepts here called ‘marginal productivity’. The marginal productivity of capital is the amount of extra output you get by adding one more unit of capital, ie \frac{dQ}{dK}. The marginal productivity of labour is the amount of extra output you get by adding one more unit of labour, ie \frac{dQ}{dL}.

So we can now say \displaystyle \frac{\partial \pi}{\partial K} = MR(MP_K)-r and \displaystyle \frac{\partial \pi}{\partial L} = MR(MP_L)-w.

Remember that to find the profit maximising point we had to set both partials to zero, ie \displaystyle \frac{\partial \pi}{\partial K} = MR(MP_K)-r=0 and \displaystyle \frac{\partial \pi}{\partial L} = MR(MP_L)-w=0 so MR(MP_K)=r and MR(MP_L)=w

So the profit maximising wage is to pay each worker the multiple of his marginal productivity multiplied by the marginal revenue. In an environment of perfect competition, P=MR so if we had perfect competition, P(MP_L)=w. This is quite an important result in micro which you use in questions to do with factor markets, because it shows that two things will drive wages up – either a rise in price of the good being produced, or an increase in marginal productivity of labour (eg workers becoming more skilled, technological advances making them more productive, or simply adding more capital so each worker has more capital to work with). Fairly logical results but that is the economics behind them.

For another example of using differentiation and algebraic manipulation, check out this little exercise in deriving the elasticity of demand for a firm in a competitive market.

Categories: Differentiation, Maths

The concept of Present Value

June 20, 2011 Leave a comment

Economists are always evaluating the value of the costs and benefits of a particular project, often over a long time scale. When you do this you don’t just add up the total sum of an expected future income stream, you have to discount it as you look into the future. This is the idea of working out the ‘present value’ of some form of return in the future.

The discount rate is basically the opportunity cost of directing resources elsewhere, so for private investment projects a good benchmark to use is the market interest rate. This is logical because it represents an alternative use of the investment funds – the firm could have just lent it to someone else (or invested it in a bank to lend it on their behalf) and earned the market rate. So if the investment is being used to finance an investment project which is forecast to earn income streams in the future, it is only worthwhile if those income streams will end up being better than what the firm could have earned by simply investing the money at the market rate of interest and taking the returns from that.

It will be easier to see with an example.

Lets say a firm is considering making an investment in a new piece of machinery which will cost £10000 now. The projections are that this piece of machinery will last for six years, and will increase the firm’s profits by £3000 in each of the first two years, by £2000 in each of the next two years, and by £1000 in the final two years before it needs to be replaced. That means in total it will bring in £12000 over 6 years.

Total value of income stream:3000+3000+2000+2000+1000+1000=12000

So it costs £10000 and brings in £12000 over 6 years…sounds like it is worth it?

Not necessarily. Suppose the market interest rate is 4%.

If you use a discount rate of 4%, then the present value of the income stream is \frac {3000}{1.04}+\frac {3000}{1.04^2}+\frac {2000}{1.04^3}+\frac {2000}{1.04^4}+\frac {1000}{1.04^5}+\frac {1000}{1.04^6} = 10758.13

Now given that the cost of the investment is 10000 now, the net present value of the project is -10000 + 10758.13 = 758.13

So yes, it is worth it. The present value of your forecast income streams is worth more than the cost of the investment, but not by the £2000 that you would get simply by adding up the values.

If the net present value of a project > 0, then it is a worthwhile investment.

When it comes to calculating the present value of a government project, the opportunity cost is the net present value of what the funds would otherwise have been used to produce. In the UK most public sector investments take 3.5% as a benchmark discount rate, this is the social time preference rate. The danger with this is that discounting can effectively disenfranchise future generations, if you are looking at a long term project that will yield gains over many years. We are still today benefiting from the investment that our ancestors made in the railways and canals etc, but if you discount every project at 3.5% then the present value of something in 100 years time is effectively the value divided by 31, and something 200 years in the future gets divided by 972. This is pretty controversial when it comes to evaluating the gains of environmental projects. So the UK government uses a declining discount rate, the 3.5% counts only for the first 30 years, and then gradually declines over different time horizons.

Categories: Geometric series, Maths