Hot answers tagged

2

The SDE can be solved similarly as in the Vasicek model. Define $F(t,r(t)) = e^{\alpha t}r(t)$, then \begin{cases} \displaystyle \frac{\partial F}{\partial t} &= \alpha e^{\alpha t} r(t) \\ \displaystyle \frac{\partial F}{\partial r(t)} &= e^{\alpha t} \\ \displaystyle \frac{\partial^2 F}{\partial r(t)^2} &= 0. \end{cases} Applying Itô's lemma ...


2

If $\mu$ and $\sigma$ are constant, $dP = \mu\; dt + \sigma\; dW(t)$ is equivalent to $P(t) = P(0) + \mu t + \sigma W(t)$ where $W$ is Brownian motion with $W(0)=0$. Then $\Delta P = P(T) - P(0) = \mu T + \sigma W(T)$ has a normal distribution with mean $\mu T$ and standard deviation $\sigma \sqrt{T}$. The density at a given $\Delta P$ is maximized with ...


2

Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$ we have $$X_t = f_{1-t}(B_t) = \frac{1}{\sqrt{2\pi (1-t)}} \exp\left\{\frac{-B_t^2}{2(1-t)}\right \}:=F(t,B_t),$$ and Itô's lemma states $$dX_t = dF(t,B_t) = \left(\frac{\partial F}{\partial t}+\frac{1}{2}\frac{\partial^2 F}{\partial B_t^2}\right)dt+ \frac{\partial F}{\partial B_t}dW_t.$$ ...


2

For $0 < t < n$ the outstanding loan balance at time $t$ computed after making the $t$-th payment is $$B^p_t = P\,a_{\overline{n-t}|j}=P\,\frac{1-v^{n-t}}{j}$$ by the prospective method. So we have that the debt ad time $t$ is $D_t=B^p_t$ and the interest payed at time $t+1$ will be $I_{t+1}=jD_t=jB^p_t$, that is $$ I_{t+1}=jB^p_{t} = ...


1

In order to find $K$ $$ L=K\,a_{\overline{5}|i}+5K\,v^5\,a_{\overline{5}|i}=K\,a_{\overline{5}|i}(1+5v^5) $$ and then $$ K=\frac{L}{a_{\overline{5}|i}(1+5v^5)}=56.88422\approx 56.88 $$ The outstanding balance at $t=4$ $$ B_4=K\,a_{\overline{5-4}|i}+5K\,v^5\,a_{\overline{5}|i}=Kv+5K\,v^5\,a_{\overline{5}|i}= 825.55 $$ The outstanding balance at $t=5$ $$ ...


1

From what I can see it's correct: $$ P_3 = P_0 \underbrace{\cdot e^{r_1} \cdot e^{r_2} \cdot e^{r_3}}_\text{one factor per year} = P_0 \cdot e^{r_1 + r_2 + r_3} = P_0 \cdot e^{0.15} \\ P_0 = P_3 \cdot e^{-0.15} \approx 8607.08€ $$


1

The cash flow is equivalent to this \begin{matrix} t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16\\ \hline A & 100 & & 120 & & 140 & & 160 & & 180 & & 200 & & 220 & & 240 & \\ B & & 100 ...


1

Here's why what you did didn't work. Let $X,Y$ be two independent normally distributed random variables with mean 2 and variance 1 (this is what you sampled from with your R code). This means that $X+Y$ is normally distributed with mean 4 and variance 2, which implies that $$ \frac{X+Y - 4}{\sqrt{2}}$$ is a standard normal random variable, so it follows that ...


1

From $\delta(t)=\frac{a'(t)}{a(t)}=\frac{\mathrm d\log a(t)}{\mathrm d t}$ we have the accumulation function $$ a(t)=\mathrm e^{\int_0^5\delta(s)\mathrm d s}=\mathrm e^{0.02\int_0^5 s\mathrm d s}=\mathrm e^{0.01\,t^2} $$ and then the future value $$ s_{\overline{n}|}=\sum_{t=1}^n a(n-t)=\sum_{t=0}^{n-1} a(t)=1+\mathrm e^{0.01}+\mathrm e^{0.04}+\mathrm ...


1

After first year, the balance is 10000*1.06-500=10100. After the second year, the balance is 10100*1.06-750=9956. After the third year, the balance is 9956*1.06-1000=9553.36. After the 4th year, 8876.5616. In general, for $P_i$ payments at the end of year $i$, the balance is $$B^r_{(n)}=10000 \times 1.06^n-\sum_{i=1}^n P_i\times1.06^{n-i}$$



Only top voted, non community-wiki answers of a minimum length are eligible