In this book, we have mainly discussed deterministic (i.e. non-random) interest,
and we will briefly introduce stochastic (i.e. random) interest,
by regarding the interest rate as a random variable. We use the following notations:
: interest rate random variable for the period
to ![{\displaystyle t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65658b7b223af9e1acc877d848888ecdb4466560)
: mean of ![{\displaystyle I_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc386d951e8ffae76357542f14e160621e6668b1)
: variance of ![{\displaystyle I_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc386d951e8ffae76357542f14e160621e6668b1)
Accumulation of single payment over several time periods
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Assume that
are independent for
.
Let
be the accumulation of a single unit sum of money invested for
years, i.e.
![{\displaystyle S_{n}=(1+I_{1})\cdots (1+I_{n}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5f40effcd02562820d57a0d9097d7cfb8e60e24)
Then, by independence,
![{\displaystyle {\begin{aligned}\mathbb {E} [S_{n}]&=(1+\mu _{1})\cdots (1+\mu _{n})\\\mathbb {E} [S_{n}^{2}]&=\mathbb {E} [(1+I_{1})^{2}\cdots (1+I_{n})^{2}]\\&=\mathbb {E} [(1+I_{1})^{2}]\cdots \mathbb {E} [(1+I_{n})^{2}]\\&=\left(\sigma _{1}^{2}+(1+\mu _{1})^{2}\right)\cdots \left(\sigma _{n}^{2}+(1+\mu _{n})^{2}\right)\\\operatorname {Var} (S_{n})&=\mathbb {E} [S_{n}^{2}]-(\mathbb {E} [S_{n}])^{2}\\&=\left(\sigma _{1}^{2}+(1+\mu _{1})^{2}\right)\cdots \left(\sigma _{n}^{2}+(1+\mu _{n})^{2}\right)-(1+\mu _{1})^{2}\cdots (1+\mu _{n})^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a3f3f9c9166d3b6bc6992e3bb08df82fa08b901)
For simplicity, further assume that
![{\displaystyle I_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc386d951e8ffae76357542f14e160621e6668b1)
's are i.i.d. (identically and independently distributed),
with mean
![{\displaystyle \mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fd47b2a39f7a7856952afec1f1db72c67af6161)
and variance
![{\displaystyle \sigma ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53a5c55e536acf250c1d3e0f754be5692b843ef5)
. Then,
![{\displaystyle {\begin{aligned}\mathbb {E} [S_{n}]&=\underbrace {(1+\mu )\cdots (1+\mu )} _{n{\text{ copies}}}=(1+\mu )^{n}\\\mathbb {E} [S_{n}^{2}]&=(\sigma ^{2}+(1+\mu )^{2})^{n}=(1+2\mu +\mu ^{2}+\sigma ^{2})^{n}\\\operatorname {Var} (S_{n})&=\mathbb {E} [S_{n}^{2}]-(\mathbb {E} [S_{n}])^{2}=(1+2\mu +\mu ^{2}+\sigma ^{2})^{n}-(1+\mu )^{2n}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ac700ef50cd7799dcb641723f87ddbf8d36b3f4)
Accumulation of investments with log-normal distribution
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If
has a normal distribution with mean
and variance
,
then
has a log-normal distribution with
parameters (not mean/variance generally)
and
. The following are some properties of random variables following log-normal distribution with parameters
and
:
- probability density function (pdf):
![{\displaystyle f(x)={\frac {1}{x\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {1}{2}}\left({\frac {\ln x-\mu }{\sigma }}\right)^{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe4f718642409dc08a02ebaa3d41af994154a03d)
- mean:
![{\displaystyle \mathbb {E} [X]=e^{\mu +{\frac {\sigma ^{2}}{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26089e466ff4f25946acbd894bb9ab14f969c021)
- variance:
![{\displaystyle \operatorname {Var} (X)=e^{2\mu +\sigma ^{2}}\left(e^{\sigma ^{2}}-1\right)=(\mathbb {E} [X])^{2}\left(e^{\sigma ^{2}}-1\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0a55cd16b94623967418e802f9010817960dd94)
Let's apply log-normal distribution to stochastic interest.
If
follows a log-normal distribution with parameters
and
, then
will be normally distributed with mean
and variance
.
Then, considering the natural logarithm of accumulation of a single investment of one unit for a period of
time units,
we have
![{\displaystyle \ln S_{n}=\ln((1+I_{1})\cdots (1+I_{n}))=\ln(1+I_{1})+\cdots +\ln(1+I_{n}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81dad1798301e04847824a2b15176c48f08307e7)
Assuming
![{\displaystyle I_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc386d951e8ffae76357542f14e160621e6668b1)
's are independent,
![{\displaystyle \ln(1+I_{t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/647601bb96f6d00610b7f011b4576d5356d5a9ea)
will also be independent.
If we further assume that
![{\displaystyle 1+I_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df1ad873668a119ec0d85d977c30c2cc1c22de99)
's are also log-normally distributed with parameters
![{\displaystyle \mu _{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8247e7beb0b3fa21c4fe35defef3a3cf93df67d)
and
![{\displaystyle \sigma _{t}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/611cb298fa3265e9e8b112f6dac2d36ff2a723e5)
,
then
![{\displaystyle \ln(1+I_{t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/647601bb96f6d00610b7f011b4576d5356d5a9ea)
's are normally distributed with mean
![{\displaystyle \mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fd47b2a39f7a7856952afec1f1db72c67af6161)
and variance
![{\displaystyle \sigma ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53a5c55e536acf250c1d3e0f754be5692b843ef5)
x,
and the
sum of independent normal random variables
![{\displaystyle \ln(1+I_{1})+\cdots +\ln(1+I_{n})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/668e73f9920c3426741f52d35371fbd276e0e5dd)
is normally distributed
with mean
and variance ![{\displaystyle \sigma _{1}^{2}+\cdots +\sigma _{n}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d30af116333ed55759162a0cb750f1a64cdfdb23)
(which is a well-known result about normal distribution).
That is,
![{\displaystyle \ln S_{n}=\ln(1+I_{1})+\cdots +\ln(1+I_{n})\sim N(\mu _{1}+\cdots +\mu _{n},\sigma _{1}^{2}+\cdots +\sigma _{n}^{2}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfde6d68d1ea2ca2cdaaeb5e176d9a17b59513b2)
Thus, if we apply log-normal distribution to stochastic interest, we can obtain this nice result (
![{\displaystyle \ln S_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/832d58fb8b1eb6139e17b5d79eaa212ddef6e990)
follows a simple normal distribution).
Example.
(a) It is given that
follows log-normal distribution with parameters
and
,
and
.
Compute
and
.
Solution: (a) Based on the given mean and variance,
we have
and
So,
![{\displaystyle {\begin{aligned}&&&{\begin{cases}e^{\mu +{\frac {\sigma ^{2}}{2}}}=1.1\\e^{2\mu +\sigma ^{2}}\left(e^{\sigma ^{2}}-1\right)=0.005\\\end{cases}}\\&\Rightarrow &1.1^{2}\left(e^{\sigma ^{2}}-1\right)&=0.005\\&\Rightarrow &\sigma ^{2}&=\ln \left(1+{\frac {0.005}{1.1^{2}}}\right)\approx 0.004124\\&\Rightarrow &e^{\mu +0.004124/2}&=1.1\\&\Rightarrow &\mu &\approx \ln(1.1)-0.004124/2\approx 0.09325.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65c30f61a9294f0828f190be5c7cdf045b0b97eb)
(b) It is further given that
is the annual yield rate for the
th year,
's are i.i.d., and follow the distribution mentioned above.
Compute the mean and variance of the accumulation of
for
years,
and the probability that its accumulated value will be less than
.
Solution: (b)
Since
,
![{\displaystyle \ln(S_{1}0)\sim N(10(0.09325)+10(0.004124))=N(0.9325,0.04124).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e40789500fb3ebfda4d404f5b7dc8942a3e502a1)
So,
![{\displaystyle S_{1}0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a9a922f2e0b20608d7ca4a908399d8f861defaa)
follows
log-normal distribution with parameters
![{\displaystyle \mu =0.9325}](https://wikimedia.org/api/rest_v1/media/math/render/svg/194c9581730b220c15336c9e876cbb740ac9d5df)
and
![{\displaystyle \sigma ^{2}=0.04124}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fa5f280cc6dba3000eee3be75f6f96ff0f0659f)
. Thus,
its mean and variance are as follows:
- mean:
![{\displaystyle \mathbb {E} [100S_{10}]=100e^{0.9325+0.04124/2}\approx 259.3790}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b60eb460e9503960f4b940e81d3c74c13b55b08)
- variance:
![{\displaystyle \operatorname {Var} (100S_{10})=100^{2}e^{2(0.9325+0.04124)}(e^{0.04124}-1)\approx 2832.5273}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db9423a702479840017a0b88e9c5e563f27b0f9c)
Then, we can compute the probability by
![{\displaystyle {\begin{aligned}\mathbb {P} (100S_{10}<220)&=\mathbb {P} (\ln S_{10}<\ln 2.2)\\&=\mathbb {P} \left(Z<{\frac {\ln 2.2-0.9325}{\sqrt {0.04124}}}\right)\qquad {\text{in which }}Z\sim N(0,1)\\&=\mathbb {P} (Z<-0.709303)\\&\approx 0.23885\qquad {\text{from standard normal table}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b684bfc6d6bdd15127622071564803c2ad5d8a9)