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Notesannuities

# Notesannuities - x = 1 g 1 r to get PV = 1 1 r 1&& 1...

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1 Annuities and Perpetuities: The goal of these notes is to derive all perpetuity and annuity formulas. It turns out that we only need to know one calculus formula: T X j =1 x j ° 1 = 1 ° x T 1 ° x (1) The proof of this calculus formula is at the end of this document (for those of you who don°t trust me!). Once we solved the hardest problem, namely a growing annuity, every other formula will directly follow (isn°t this great?). You should be really excited now! A growing annuity has the following cash ±ows: t = 0 t = 1 t = 2 t = 3 ... T 0 1 1 + g (1 + g ) 2 ... (1 + g ) T ° 1 (2) Then, we can conveniently write this as: (The P sign just indicates summation) PV = T X t =1 (1 + g ) t ° 1 (1 + r ) t (3) First trick: Multiply and divide by (1 + g ) : PV = T X t =1 (1 + g ) t ° 1 (1 + r ) t 1 + g 1 + g = 1 1 + g T X t =1 (1 + g ) t (1 + r ) t (4) = 1 1 + g T X t =1 ° 1 + g 1 + r ± t Second trick: De²ne x ± 1+ g 1+ r so that: PV = 1 1 + g T X t =1 x t (5) We can almost use the formula in equation 1 (look at the exponent!!). Mr. Smart has another idea: Multiply and divide by x : PV = 1 1 + g T X t =1 x t x x (6) = x 1 + g T X t =1 x t ° 1 (7) 1

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Now we can apply formula 1 for the term: P T t =1 x t ° 1 = 1 ° x T 1 ° x and obtain: PV = x 1 + g 1 ° x T 1 ° x (8) The last thing we need to do is to plug in the de²nition for
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Unformatted text preview: x = 1+ g 1+ r to get: PV = 1 1 + r 1 & & 1+ g 1+ r ± T 1 & 1+ g 1+ r = 1 & & 1+ g 1+ r ± T (1 + r ) h 1 & 1+ g 1+ r i = 1 & & 1+ g 1+ r ± T 1 + r & (1 + g ) = 1 & & 1+ g 1+ r ± T r & g (9) That±s it!! From here, we can easily derive all formulas: T & period Annuity (no problem, set g = 0) 1 & ( 1 1+ r ) T r T & period growing annuinty (we solved for this!) 1 & ( 1+ g 1+ r ) T r & g Growing Perpetuity (let T ! 1 ), works only if g < r , why?? 1 r & g Regular Annuity 1 r Proof. T X j =1 x j & 1 = T & 1 X j =0 x j = ² 1 + x + x 2 + x T & 1 ³ Multiply both sides by (1 & x ) and you obtain: (1 & x ) T X j =1 x j & 1 = (1 & x ) ² 1 + x + x 2 + x T & 1 ³ = ² 1 + x + x 2 + x T & 1 ³ & x ² 1 + x + x 2 + x T & 1 ³ = ² 1 + x + x 2 + x T & 1 ³ & ² x + x 2 + ::: + x T ³ = 1 & x T Hence: (1 & x ) T X j =1 x j & 1 = 1 & x T or T X j =1 x j & 1 = 1 & x T 1 & x 2...
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