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5

Recall from Geometric series, we have $$\sum_{t=0}^{\infty} x^t = \dfrac1{1-x}$$ Differentiating both sides, we obtain $$\sum_{t=0}^{\infty} tx^{t-1} = \dfrac1{(1-x)^2} \implies \sum_{t=0}^{\infty} tx^{t} = \dfrac{x}{(1-x)^2}$$ Plugging in $x=\frac1{1+y}$, we obtain what you want.


2

Suppose you need to pay a net of $N$ to the business. Let the actual amount you give be $A$. We know $N$ and want to find $A$. The processing fee is $(0.024)A +0.24$. In addition, VAT has to be paid on that, an amount equal to $(0.23)\left[(0.024)A+0.24\right]$. So the total lost from $A$ to fee and VAT is $(1.23)\left[(0.024)A+0.24\right]$. It follows ...


2

$$dV = -\beta Vdt +\sigma dB, \space V(0)=0$$ Treat this like an integrating factor ODE with initial condition $V(0) =0$. Define $Y=e^{\beta t}V$, using Ito: 1)$$dY = \beta e^{\beta t}Vdt + e^{\beta t}dV = \beta e^{\beta t}Vdt - \beta e^{\beta t}Vdt + e^{\beta t}cdB = e^{\beta t}cdB$$ 2) $$\Rightarrow Y(t) = Y(0) + \sigma \int_{0}^te^{\beta s}dB ...


2

We have that $$ V = W_t^2 - t$$ hence (written with $t$ and $x$, where we plug in $S_t = W_t$ for $x$), we have $$ V(x,t) = x^2 - t$$ Now \begin{align*} \frac{\partial V}{\partial t} &= -1\\ \frac{\partial V}{\partial x} &= 2x\\ \frac{\partial^2 V}{\partial x^2} &= 2 \end{align*} Hence (note that $S_t = W_t$) \begin{align*} ...


1

Simply because adding $0.2x$ to $x$, giving you $1.2x$, is not the same thing as $x/0.8$. This is because $$1/0.8 = 1.25$$ You're not increasing by the same factor, the second option would be $1.25x$ instead of the very correct $1.2x$.


1

$e^{\frac{m(ln(1+\frac{R_m}{m}))}{m}} \rightarrow e^{ln(1+\frac{R_m}{m})} \rightarrow 1+\frac{R_m}{m}$ Plugging it back into the larger equation $m(1+\frac{R_m}{m}-1) \rightarrow R_m$


1

Well, when it's known what is constant and what it's not, the key to solving this equation is observation that it's first order inhomogeneous linear equation: $$ \frac{d\omega}{dt} + c \left (\frac{be^{-bt}}{ae^{-bt}-C_1} \right) \cdot \omega = f(t) $$ General solution of this ODE is a linear combination of any solution of it and general solution of ...


1

\begin{align*} f & = S_0\left(\frac{f_u - f_d}{S_0u - S_0d}\right)\left(1 - ue^{-rT}\right) + f_ue^{-rT}\\ & = S_0\left(\frac{f_u - f_d}{S_0(u - d)}\right)\left(1 - ue^{-rT}\right) + f_ue^{-rT}\\ & = \left(\frac{f_u - f_d}{u - d}\right)\left(1 - ue^{-rT}\right) + f_ue^{-rT}\\ & = \frac{\left(f_u - f_d\right)\left(1 - ue^{-rT}\right) + ...


1

I DID IT! $$f = S_0\left(\frac{f_u - f_d}{S_0u - S_0d}\right)\left(1 - ue^{-rT}\right) + f_ue^{-rT}$$ $$f = \frac{S_0f_u - S_0f_d}{S_0u - S_0d}\left(1 - ue^{-rT}\right) + f_ue^{-rT}$$ $$f = \frac{S_0\left(f_u - f_d\right)}{S_0\left(u - d\right)}\left(1 - ue^{-rT}\right) + f_ue^{-rT}$$ $$f = \frac{f_u - f_d}{u - d}\left(1 - ue^{-rT}\right) + f_ue^{-rT}$$ $$f ...


1

It's quite common to distinguish between Type A arbitrage and Type B arbitrage. We say that a trading strategy is a type A arbitrage if it has a positive initial cashflow and no risk of future loss. type B arbitrage if it has a nonnegative initial cashflow, no risk of future loss and a positive probability of future profit. I think what you are referring ...


1

Set $X_t = W_t^2 + at + b\int_0^t W_s^2\, ds$, so that $$dX_t = (1 + a + bW_t^2)\, dt + 2W_t\, dW_t$$ and $$d[X,X]_t = 4W_t^2\, dt.$$ Using Ito's lemma, we get $$d[e^{X_t}] = e^{X_t}\, dX_t + \frac{1}{2} e^{X_t} d[X, X]_t = e^{X_t}\bigl((1 + a + (b + 2)W_t^2)\,dt + 2W_t\, dW_t\bigr).$$ For $e^{X_t}$ to be a martingale, the drift term must vanish, ...



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