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A series of payments made at fixed points in time are called annuities. Essentially, you are summing up a finite geometric series. If you want to calculate how much you'd have to pay per month to pay off a $\$ 4773$debt at an annual effective rate of$20 \%$, we'd first convert our annual effective interest rate to the effective interest rate per month: ... 2 Let$B_1$be your current balance that you will start paying this month,$B_i$for month$i$, and$P$the amount of the fixed payment. You'll pay off your bill in month 12, and want the 13th statement to be 0. $$B_1 = B_1$$ $$B_2 = \left(1+\dfrac{0.2}{12}\right)(B_1-P)$$ $$B_3 = ... 2 Are you sure you're getting paid interest at all? The difference between any two consecutive numbers in the sequence you give is 321.63. This would seem to indicate that you're putting in 321.63 per month and getting no interest - if you were getting interest (even some very small amount) you'd expect the difference between the month n and month n + 1 ... 2 I'll show (i). After the first month, the amount owed, y_1, is$$y_1 = x_n \left (1+\frac{r}{12}\right ) - my_2 = y_1 \left (1+\frac{r}{12}\right ) - m = x_n \left (1+\frac{r}{12}\right )^2 - m \left (1+\frac{r}{12}\right ) - my_3 = y_2 \left (1+\frac{r}{12}\right ) - m = x_n \left (1+\frac{r}{12}\right )^3 - m \left [ \left ... 1 You are rarely asked to find the variance of a weighted distribution without knowledge of the joint distribution of A and B. If only given two distributions without any indication of their correlation, it is not possible to calculate$p$. In particular, I dont know how you got the$Cov(A,B) = ((0.12−.1)(0.08−.1))$part. Check the question and the solution ... 1 So let me add how to get to the estimate formula: As usual, you start paying at the end of the first month, and as in Tylers answer you get $$B = P⋅\frac{1 - (1+i^{month})^{-12}}{i^{month}}$$ however, as in David Peterson's answer, banks compute$i^{month}=i^{annual}/12\$. Now cite Sir Isaac Newton for the binomial series $$... 1 For a single event of 30\% inflation, after one year the price goes from 160,000 to 208,000 as you say. If the inflation is 2.5\% per month and you apply it every month, in one year the 160,000 goes to 160,000(1.025)^{12}\approx 215182.21 The difference represents the effect of compound interest. Normally annual inflation should take account ... 1 I'm assuming this is an assignment question so I'm just going to try to point you in the right direction (and only on the first part: you said in a comment that this was what you wanted help with). Do any decision makers with (increasing) utility function agree about preferring risk X_1 to X_2? The expectations of X_1 and X_2 are equal: ... 1 It is much easier if you consider the value of each deposit separately and add them up. It is 3000(1.1+1.1^2+\dots +1.1^{10}), assuming you make the deposit at the beginning of the year and the withdrawal at the end of year 10. The part in the parentheses is a geometric progression which can be summed, giving a final value of ... 1 Answer: Here is a slick way to calculate the Loan Amount and Principal and Interest for any year without using great formulas. Part I: The Loan amount is 1513.67 Part II: Beg Bal and End Bal and the interest , principal schedule for all years. Basically this is an amortization schedule just like what the other responder alluded to. I am sorry, I ... 1 The excel is wrong. Keep in mind that the payment first goes towards interest for the year - then towards repayment of the principal. A more theoretic approach of solving it: Consider the loan amount as the sum of the present value of two different annuities. I will be using standard annuity notation. B_0 = 100a_{12|0.05} + 10Da_{12|0.05} Which gives ... 1 I hope this can get you started: What I understood from the picture is that the person who wants to be insured has an exponential utility function with parameter one$$u(x)=-e^{-x},$$and that the risk X is random with distribution \text{Exp}(2), that is, the expected value of a function f of X is$$\mathbb E(f(X))=\int_0^\infty f(x)\,2e^{-2x}\, ...