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3

If the power is the loop, then the $1+$ hides in the += assignment. Expand it to P = P + interest or further to P = P + P * interestRate / 100 and finally simplify to P = P * (1 + interestRate / 100) and you are there.


2

Answer: $$a_{(2-4)} = \frac{a(4)}{a(2)}$$ $$A(2) = A(1).\frac{a(2)}{a(1)}$$ $$A(2) =240\times\frac{1.5}{1.2} = 300$$ $$A(4) = 300\times\frac{3}{1.5} = 600$$ $$A'(4) = A'(3)\frac{a(4)}{a(3)}$$ $$A'(4) = 300\times\frac{3}{2}$$ $$A'(4) = 450$$ $$I_1 = A(4)-A(2) = 600-300 = 300$$ $$I_2 = A'(4) - A'(3) = 450-300 = 150$$ $$I = I_1 +I_2$$ The required ...


2

I agree that @BCLC is right on saying that I have used risk neutral information The Edited Answer is $ V(0) = 15\times90+ 10\times25 = 1600$ Now compute V(T) $$V(T) = 1800, \text{ if stock goes up}$$ $$1800 = 30\times 10 + 100\times 15$$ $$V(T) = 1700, \text{ if stock goes down}$$ $$1700 = 20\times 10 + 100\times 15$$ $V(T) = 15\times A(T) + ...


2

Under the time value of money, due to the constant interest rate, your cash flow stream is equivalent to $50$ for $20$ years, $50$ from years $2$ through $20$, and $100$ from years $3$ through $20$. (Draw a timeline of both scenarios to convince yourself that this is true.) The $50$ for $20$ years has present value ...


1

12% per annum compounded every 6 months is 3 payments of 6%. $15000\cdot (1+0.06)^3=17865.24$ So $17865.24-15000.00 = 2865.24$


1

The payments are made continuously in the second case. So you really aren't evaluating "at $t=m$", they are just your integral bounds. You can think of the continuous payments beginning instantaneously after time $t=m$ and ending instantaneously before time $t=m+n$ if that helps.


1

The notation is a bit confusing for annuities (and not something I've seen before). But to understand this, you need to recall that $a_{\overline{n}|}$, the present value of an annuity-immediate, finds the present value of $n$ payments of $1$ starting at time $1$ to time $n$. The payment at time $0$ is excluded. This is true no matter at what time you ...


1

Let $i$ be the interest rate. $$X=\left ( \frac{1}{1+i}+\frac{1}{(1+i)^2}+\frac{1}{(1+i)^3} \right )\sum_{k=1}^{13}\frac{k}{(1+i)^{39-3k}} $$ The sum is in the form $$ \sum_{k=1}^{n}kz^k=z\frac{1-(n+1)z^n+nz^{n+1}}{(1-z)^n} $$ so $$ ...


1

You are almost there. Try doing a change of variables in the exponential part of the integral. Make it so that you get "$e^{-\frac{1}{2} x^2}$". Then you can relate this to the cdf of the normal distribution, ie $$ \Phi(z) = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} y^2} dy. $$ Note also that, since the integrand is a pdf, $$ 1 - \Phi(z) = ...


1

I'm not familiar with all the financial formulae, but the concepts are simple. Simple interest is always calculated on the original principal. Interest per year is $\frac{$11520}{2} = \$5760$, giving an annual interest rate of $\frac{5760}{24000} \times 100\% = 24 \%$. Assuming that's compounded annually, in two years you would have $1.24^2$ of your ...


1

For simple interest, $A(2)=(24000+11520)=24000(1+2i)$ We get $i=0.24$ For compound interest, $A(2)=24000(1+0.24)^2=36902.4$ $I_{[2,0]}=A(2)-A(0)=36902.4-24000=12902.40$


1

The formula used isn't right. For a principal $P$, rate of interest $r$%, time in years $n$, the amount $A$ when compounded annually is given by $$A=P(1+\frac{r}{100})^n$$ Thus $$31104=P(1+\frac{20}{100})^3$$ $$31104=P(\frac{6}{5})^3$$ $$P=\frac{31104*125}{216}=18000$$ The principal must be $18000$.


1

Due to the upfront fee your investment is smaller than what you've effectively paid thus lowering your participation from the 25 % raise. Furthermore the redemption fee also lowers your profit. Taking both into account you paid \$4.000 for 910 shares of the fund (due to the 9 % fee). At the end these are worth \$ 4.550 before fees. Subtracting 1.5 % leaves ...


1

In the context it is clear that $A_R^T = \frac 1{1 + R} + \frac 1{(1+R)^2}+ ...\frac 1{(1+R)^T}$. Which by geometric series = $\frac {1-(1+R)^{T+1}}{R}$ So I think in this case it is simply a double index. There are two values that determine $A$; they are $R$ and $T$ so we need two indexes to reference $A$. We could have used $A_{R,T}$ just as easily. ...


1

For your first question, you should not mix $df(t, T)|_{T=t}$ with $df(t, t)$. For $df(t, T)|_{T=t}$, there is no differential to the second argument, that is, when taking the differential, the second argument is ignored, while setting $T$ to $t$ after the differential is done. However, for $df(t, t)$, you need to consider the differential for both the ...


1

You are fine for 1). For 2), if you bet a fraction $K$ of your account, then on a win your account is multiplied by (what?). On a loss your account is multiplied by (what?). It doesn't matter the order of wins or losses, just the number of each. Each one becomes an exponent. Your expression does not allow for the fact that your account balance, and ...


1

The idea Let $A_n$ the amount you have at the $n$-th year with $A_0=500$. Each year, you get $0.025\cdot A_n$ more than what you had previously (i.e. $A_n$). So, each year you'll have $$A_{n+1}=A_n+0.025A_n=1.025A_n=...=1.025^{n+1}A_0.$$ You now have all the elements to solve your exercise.


1

For a proof by contradiction, assume simple interest: $A(5)=A(0)(1+5i)$ $A(4)=A(0)(1+4i)$ $A(10)=A(0)(1+10i)$ $A(9)=A(0)(1+9i)$ $i_5=\frac{A(5)}{A(4)}-1$ $i_5=\frac{A(0)(1+5i)}{A(0)(1+4i)}-1=i_5=\frac{(1+5i)}{(1+4i)}-1$ $i_10=\frac{A(0)(1+10i)}{A(0)(1+9i)}-1=i_{10}=\frac{(1+10i)}{(1+9i)}-1$ Solving for $i_5=i_{10}, ...


1

If you manipulate the expression you get $$P = A\dfrac{\left[(1+i)^n - 1\right]}{i(1+i)^n}$$ $$(1+i)^n(\frac{Pi}{A}) = (1+i)^n-1$$ $$(1+i)^n \left[1-\frac{Pi}{A}\right] = 1$$ $$(1+i)^n = \dfrac{1}{\left[1-\frac{Pi}{A}\right]}$$ Taing log on both sides We get $$n log(1+i) = log\left(\dfrac{1}{\left[1-\frac{Pi}{A}\right]}\right)$$ $$n = ...


1

Let´s assume a simple interest rate of r and an initial amount of 1 monetary unit. And each period the account is increasing by 1 monetary unit. After the first period you get an interest of $r$. After two periods you get an interest of $r(1+1)=2r$ After $(t-1)$ periods you get an interest of $(t-1)\cdot r$ After n periods you get an interest of $n\cdot r$ ...


1

Answer: $I_{(0,t)} = A(t) - A(0) $ We also know that: $I_{(0,1)} = A(1) - A(0) = 1$ $I_{(1,2)} = A(2) - A(1) = 2$ $I_{(2,3)} = A(3) - A(2) = 3$ ... $I_{(t-2,t-1)} = A(t-1) - A(t-2) = t-1$ $I_{(t-1,t)} = A(t) - A(t-1) = t$ Summing the above, you will see mass cancellation and the sum S reduces to $S = A(t) - A(0) = I_{(0,t)} = (1+2+3+\cdots|t) = ...


1

Answer: The correct alternatives are (1) and (2): $ [\underline{81,062.5} + 19,062.5(1+0.04)^{1} -95,000] (1 + 0.04)^{2}\tag{1}$ $ [\boxed{87000} + 19,062.5(1+0.04)^{1} -95,000(1+{\boxed{0.0625}})^{1}] (1 + 0.04)^{2}\tag{2}$ This is because, you are adding the interest for the third year in the second solution where as you are subtracting the interest ...


1

Answer: Let $i_2$ is the interest rate for $t_2$. and let $i_1$ be the interest rate for time $t_1$. Let the forward rate be $r$. From the diagram and simple financial basis, it is true that $$(1+r)^{(t_2-t_1)} (1+i_1)^{t_1} = (1+i_2)^{t_2}\tag{1}$$ $$v(t) = \frac{1}{(1+i)^t}$$ Thus $$v(t_1) = \frac{1}{(1+i_1)^{t_1}}$$ $$v(t_2) = ...



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