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4

Refine your initial formula in the following manner. $$A_1=A_0+\frac rnA_0+r(A_0+\frac rnA_0)+\frac rn(A_0+\frac rnA_0+\frac rn(A_0+\frac rnA_0))+\dots$$ where we repeat this $n$ times. This is interest compounded $n$ times in one year. In words, to make it less confusing, we start with $A_0$. $$A_0$$ We add on $\frac rn$ times this initial amount ...


3

Those are two different things, in the following sense: Change of measure is a mathematical operation, useful in various ways and situations. Change of numéraire has an economic rationale; instead of using one numéraire, you use another one. To make things look nicer you can change your measure at this point (mainly because under the new measure the ...


2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...


1

TLDR: degenerate is short for degenerate elliptic. Consider a second-order differential operator $$\mathcal{L}u\equiv \sum_{i,j=1}^n a_{ij} u_{ij}+\sum_{i=1}^n b_i u_i + cu$$ where I am using subscripts on $u$ to denote derivatives. In your case, $$\mathcal{L}V\equiv rSV_{S}+\frac{1}{2}\sigma^{2}S^{2}V_{SS}-rV.$$ $\mathcal{L}$ is said to be elliptic at a ...


1

The first formula you gave is simplified and assumes that the interest is compounded once per period. The bottom formula is more general and does not make this assumption. They are both correct though, just different assumptions.


1

I will just point out that $$(1+r)^{1/n}=\sum_{k=0}^{\infty}\binom{1/n}{k}r^k=1+\frac{r}{n}+\cdots$$ by using the Taylor series, where $\binom{x}{k}=\frac{x(x-1)\cdots (x-k+1)}{k!}$ when $x$ is real. So letting $r_n$ being the rate of the $n$-th part of some unit of time rate $r$, we have $$r_n=(1+r)^{1/n}-1\sim\frac{r}{n}$$ by ignoring terms of higher ...


1

In the formula $A_t=A_0(1+r)^t$, correctly interpreted, $r$ is the interest rate per compounding period and $t$ is the number of periods. Nowhere in the formula is there any definition of what the compounding period is--it could be an entire year, or a month, or an hour. People sometimes write $A_t=A_0\left(1+\frac rn\right)^t$ instead when there are $n$ ...


1

You are right. Let the (yearly) interest rate denoted as $i$. Then the annual equivalent interest rate (AER), compunded m times a year, is $i^e_m=\sqrt[m]{1+i}-1$. You are right that there is a difference to the annual percentage year rate (APR) which is $i^p_m=\frac{i}{m}$. If the bank pays interest more than once in a year ($m>1$) then it is an ...


1

Roughly and intuitively speaking, a self financing strategy is (for me) a trading strategy which requires no extra cost during the trading except for the initial capital. Suppose that we are in a continuous time case. Let $(\Omega,\mathcal{F},P)$ be a prob space with filtration $\mathbb{F}$, $S$ (a semimartingle) be the discounted price process of the stock, ...


1

Hints: The future value is $FV=100\cdot q \cdot \frac{q^{60}-1}{q-1}$ with $q=1+\frac{0.12}{4}=1+0.03$ The present value is $PV=\frac{FV}{q^{60}}$


1

This is very basic binomial probability. Just take the very first formular on this wikipediaarticle: click me Where p would be your 55% and k the 1,2...,11,12 and n=12


1

Let $P$ be the principal, and $r$ be the rate of interest in percentage. Then we have, $$\tag13920=P\left(1+\frac{r}{100}\right)^1$$ $$\tag24390.4=P\left(1+\frac{r}{100}\right)^2$$ Squaring $(1)$ and dividing by $(2)$, $$P=\frac{(3920)^2}{4390.4}$$


1

The six month calculation assumes compounding at six month intervals. The fair way to compare it with the year calculation would be to divide by $\sqrt {1.1}$ for the payment at six months, so the present value would be $\frac {50}{\sqrt {1.1}} + \frac {50}{1.1}\approx 93.13$ This reflects the fact that the proper interest for six months to equal $10\%$ per ...


1

If we have continuous compounding the future value after one year is $C\cdot e^r$ For $n$ yearly payments we have the series $FV=C+Ce^r+Ce^{2r}+Ce^{3r}+\ldots+Ce^{(n-1)r}$ Using the closed form of a geometric series we get $FV=C\cdot \frac{e^{rn}-1}{e^r-1}$ To get the present value the FV has to be discounted $n$ times $PV=C\cdot\frac{1}{e^{rn}}\cdot \...


1

Let $ X_t=aB(t)-t$ and $Y_t=\exp(2B_t-2t)$, we have $$dX_t=-dt+adB_t$$ and $$dY_t=\underbrace{\left(-2e^{2B_t-2t}+\frac{1}{2}(2)^2e^{2B_t-2t}\right)}_{0}dt+2\,e^{2B_t-2t}\,dB_t=2\,e^{2B_t-2t}\,dB_t$$ $$d(X_tY_t)=Y_tdX_t+X_tdY_t+d[X_t,Y_t]$$ as a result $$d(X_tY_t)=e^{2B_t-2t}(-dt+adB_t)+(aB(t)-t)2e^{2B_t-2t}\,dB_t+2a\,e^{2B_t-2t}dt$$ thus $$d(X_tY_t)=(-1+2a)...



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