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6

Day-count conventions. You can't live with them, you can't live without them. The reason the prices differ is that the pricing engine can't calculate correctly the time over which the first coupon is discounted, and thus it gets slightly different discount factors to apply to the coupon amounts. Please sit down, it'll take some explaining. Ultimately, both ...


6

I'm familiar with the library, but not with the way it is exported to R. Anyway: gearings are optional multipliers of the LIBOR fixing (some bonds might pay, for instance, 0.8 times the LIBOR) and spreads are the added spreads. In your case, the gearing is 1 and the spread is 0.0140 (that is, 140 bps; rates and spread must be expressed in decimal form). ...


6

This is called on the run/off the run arbitrage, a type of convergence trade. The basic idea is that as the liquidity premium disappears for the on-the-run issue, the price will fall and converge to the price of previous issues. Here are a couple papers - http://people.stern.nyu.edu/lpederse/courses/LAP/papers/SearchBargaining/VayanosWeill.pdf ...


5

As John already mentioned the formula for calculating yield to maturity is independent of any risk-related numbers. Its just the connection between coupons, time and price. In theory, default-risk can be seen as already incorporated in the yield. The yield spread between the bond and a comparable investment without default risk is a measure for the default ...


5

The general idea is to bootstrap the discount factors in the correct order, based on the data you have given. I'm going to make some assumptions that your bonds are paying annual coupons. The longest maturity is 2.5 years, meaning you need discount factors for 6M, 1.5Y and 2.5Y. The 6M deposit has a rate of 5%, this tells you that you should use the 5% rate ...


4

Note that $\frac{F(0,s,T)}{F(0,t,T)} = \frac{T-t}{T-s}\frac{B(0,s)-B(0,T)}{B(0,t)-B(0,T)}$ and $\frac{F(s,s,T)}{F(s,t,T)} = \frac{T-t}{T-s}\frac{B(s,s)-B(s,T)}{B(s,t)-B(s,T)}$. Multiplying the numerator and denominator of the last expression with $B(0,s)$ and noting that $B(0,s)B(s,u)=B(0,u)$ (investing one Dollar for $s$ years and then for another $u-s$ ...


4

To add to emcor's answer, if a bond defaults, you do not automatically get the "recovery" amount immediately, you get some unknown amount at some unknown time in the future, possibly years later, and greatly depending on your particular bond's covenants and seniority. If you are trying to consistently price bonds, you might be better off implying the ...


4

you can view a bond as a floating rate note plus a swap from floating to fixed. Floating rate notes are always at par after coupon payments (ignoring credit risk...) so the pricing of a bond is the same as that of a swap. So the pricing of a callable bond is the same as that of a cancellable swap. A cancellable swap can be viewed as a swap minus the ...


3

The intuition behind Macaulay Duration is the average time it takes to get all the cash flows from a bond. Think of it as computing the centre of gravity for a see-saw. You can find the image depicting the same here: This should immediately tell you that Macaulay Duration for Zero coupon bond is the maturity of the bond. In continuous discounting ...


3

Your overall approach is correct. However to my knowledge it is formally more appealing to work with a parameterized and smoothed yield curve. Basically one assumes that the yield curve can be described by a smooth function $r(t,\alpha, \beta,\gamma)$ (mostly of three parameters) Given a set of market data $Y(t,T_1)\dots Y(t, T_n)$ one looks for ...


3

Bond Price Dynamics I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those. Due to the model being affine the bond price is given by $$P(t,T)=A(t,T)e^{-r(t)B(t,T)}$$ you can find the exact formulas for $A(t,T)$ and $B(t,T)$ in this document (or just read ...


3

I would answer your question with no. First: what do you need the risk free rate for? If you want to price equity derivatives then probably a short money market rate would better fit this purpose. Second: the maturity. Look at yield curves. The short end is usually at a very different level than the 10 year rate. So two times no. A small "no" for ...


3

Dirty bond price refers to the price of a bond that reflects the interest that has accrued since the issuance of the bond or last coupon payment. It has nothing to do with how you discount cash flows but just whether accrued interest is priced in or not. Thus, dirty and clean bond prices apply to all bonds that pay intermittent cash flows.


3

Despite seeing one of the answers as having been chosen as the desired one, I like to offer a different perspective: Whether the yield to maturity can be derived from a bond's price in a rather identical fashion, regardless of the inherent risks is, imho, not the point of the OP, given I understood the question correctly. The yield of a bond with risk ...


3

I haven't read Yue-Kuen Kwok's book, so it's hard for me to comment on it. Based on my personal experience, I'd recommend the following literature, depending on what you're trying to accomplish: If you're on the quant-path, I think a lot of practitioners would recommend Interest Rate Models – Theory and Prctice (Damiano Brigo & Fabio Mercurio): This ...


3

I think what you wrote is correct. I'll rephrase everything according to my way to give you another point of view. The price of a coupon bond at time $t = 0$ is the sum of the discounted cashflows given by the coupons and the face value: $$ P_0 = F \cdot D(0, T_n) + \sum_{i=1}^{n} 11.04\% \cdot 0.5 \cdot F \cdot D(0, T_i) $$ where $F$ is the face value, ...


2

The way you are trying to solve these equations makes assumptions about the rates less than 10 years and therefore the shape of the yield curve. \$90 is the value of 8% coupons plus a 10-year zero-coupon bond. \$80 is the value of the 4% coupons plus a 10-year zero-coupon bond. 8% coupons are worth twice 4% coupons over the same period, regardless of the ...


2

The value it is giving you is incorrect. This is known because every option-adjusted spread calculation in existence is incorrect. I am joking here, but only a little bit. They really are all terrible. In any case, there do exist different types of OAS calculations, so you have to know which stochastic model this external utility claims to be using. ...


2

The conversion factor associated with each bond the futures' delivery basket is constructed such that the invoice prices of the bonds are identical under the assumption that the yield curve is flat at the level of the futures' notional coupon. Therefore, the bond with the highest duration will be the CTD when yields are above the notional coupon and the bond ...


2

First, I am not sure which exact statement was made. Also, you cannot just say "without CF" because you are essentially creating an artificial market with messed-up utility. In summary the cheapest-to-deliver bond is: The bond that results in the smallest loss or greatest profit for the futures seller. Futures sellers have to buy the bonds they are going ...


2

This is wrong: effectiveDate / Valuation_date = 10 May 2014 Good that you included the ISIN, which states that the effective date (as contrasted with the issue date) was a few days after 03 May 2013.


2

Let's approximate the time to maturity to be 3 years and 10 months. Assume that coupon is paid on March 6 each year. Let face value $F=100$ and coupon $c=0.07375F$. Let the discount factor be $d(0,T)=e^{−r T}$ where $r=0.06535$. The price of the bond is $$ce^{−10/12 \bullet r}+ce^{−22/12 \bullet r}+ce^{−34/12 \bullet r}+(F+c)e^{−46/12 \bullet r}=103.24 \; ...


2

Normally, you do indeed add a credit spread $s$ to the risk-free spreads to price the bond. That is, if the coupons are $c_i$ at times $t_i$ and the notional is $Y$ then you price it as $$ R\!B(t) =Y \exp{\left( -\int_t^T s(x)+r(x) dx \right) } +\sum_{i \ni t_i>t}^{N_c} c_i \exp{\left( -\int_t^{t_i} s(x)+r(x) dx \right) } $$ Normally you have too ...


2

Like Aksakal already mentioned in his comment it might depend on the duration formula you use. (see e.g. the wikipedia page or here) It can also depend on the type of instrument as mentioned by Richard. This topic has also been already discussed on the Wilmott Forum (their proposed solution is a reverse floater) Theoretically bonds with embedded options ...


2

I think the problem here is that the recovery rate is not a fixed parameter, it is only estimated from past defaults. You never know what the actual recovery rate will be, so markets change their view all the time. If your estimated recovery rate is higher than the bond price, you can either assume that the bond is underpriced, or your estimated recovery ...


2

In this context, I believe carry refers to the sum of "pure" carry + roll down. Carry, in the most general sense, is the return of a position in a static world; i.e., assuming time is the only variable that is changing, what's your holding period return on a trade? When you buy a bond, the "total carry" is the sum of 1) "Pure" carry – you get interest ...


2

It is useful in risk reports because it tells a trader the interest rate risk of each bond in his portfolio. It is true that it is based on an infinitessimal yield curve bump but the difference between this and 1 basis point bump is usually very small and is considered negligible by many. Note that more sophisticated traders do also calculate the dollar ...


2

The FRA A FRA is an agreement to exchange cash flows; the FRA in question is: Start 15/9/14 End 15/5/15 which is 242 days. USD Money Market quoting is Actual/360, so the accrual factor here is 242/360 = 0.6722. The FRA cashflows, therefore, are: on 15/9/14, Fix pays $\$1m * (0.6722 * 0.05) = \$33,611.11$, and Float pays $\$1m * (0.6722 * L)$, where L ...


2

It is a Wiener integral as your integrand is a deterministic function of time. It is known that the Wiener integral is stationary gaussian process with independent increments. So $z(t) \sim \mathcal N\left(0, \int_0^te^{-2k(t-s) }~ds\right)$ and $(z(t)-z(s)) \amalg z(u), \ \forall u,s,t \in \mathbb R_+ \text{ such that }u\leq s, s\leq t $ or alternatively ...


1

For the sake of completeness: Taking pbr142's comment into account and working in the setting you described. Set $f(C,y)=Ce^{-y} + 2Ce^{-2y} + 3Ce^{-3*y} + 300e^{-3y}$. Write $B(C,y)$ instead of $B$. Applying the quotient rule to $\frac{\partial D(C,y)}{\partial C}$ with $D(C,y)=\frac{f(C,y)}{B(C,y)}$. This leads to the following expression ...



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