If you are computing FVA, presumably you are trying to account for shareholder value only (and not 'whole bank' = shareholder + bondholder value). The correct discount factor is then OIS + funding spread + counterparty credit spread. Search for Mats Kjaer's 'Consistent XVA Metrics' papers on SSRN.

<t>I can't see this can be used to get an upper bound. The bank generally trades a cancellable swap, paying a structured leg, receiving a funding leg. The bank retains the option to cancel this. In effect, the bank are long an option to enter into a swap to receive structured coupons and pay funding...

<t>QuoteDo you mean that when you do not know the convenience yield of the collateral (the EUR cash) you use by default the collateralisation in cash, in the currency of the contract?I don't fully understand your question I am afraid. I'm trying to describe how one should build a set of curves for v...

<t>It is easiest here to totally forget about pricing uncollateralised trades, and concentrate on building curves consistent with the relevant collateral arrangement. In the scenario you have described, it is typical to use the xccy basis swap rates to build the curve used to discount USD cashflows ...

<t>When you change numeraire you are basically switching the probability measure in which you take expectations. Say you price some contract [$]C[$] with payoff [$]f(S_T)[$] in the risk neautral measure [$]\mathbb{Q}[$] where [$]B_t=e^{-rt}[$] is your numeraire:[$]C(t)/B(t)=\mathbb{E}_t[\frac{f(S_T)...

yes, messed up the probs on my tree (somehow I morphed into a 3 tile game after the first attempt...). If you can miss, my instinct was it would take twice as long on average, but I did not get twice the initial result. Now to check with correct probs.

I also get four. And just for fun: what about if there is an even chance of missing all the squares; conditional on not missing, all squares are equally likely.

<t>A Note on Construction of Multiple Swap Curves with and without Collateral; Fujii, Shimada, Takahashi. Search SSRN for it. Also other papers by these three. I'm not a big fan of the way they set up their theoretical framework (in my opinion Cooking with Collateral by Piterbarg is much more sound;...

<t>On your first question about 'justifying OIS discounting', the best argument I have found is Piterbarg's 'Cooking with Collateral' in Risk, where he explicitly constructs an arbitrage that exists if you don't discount at the rate paid on the collateral. It seems to me that a lot of the other just...

QuoteOriginally posted by: CrashedMintyou should normalize the scale to 0-10, with "very unhappy" mapping to 0, "neither" mapping to 5 and "super happy" mapping to 10, then call it the CONNOR HAPPINESS INDEX(tm). You're welcome.Can I short it? Would be an excellent hedge...

<t>Hmmm, maybe I'm actually trying to answer the lasso version? If you can tie a tight knot, I could just tie the rope around a circular cross section of the cone (even though this isn't a geodesic). But so long as the rope is above me, I will be able to pull myself up.If the rope is around a geodes...

<t>Rumour has it that an esteemed member of TCM at Cambridge used to ask this question to potential PhD students, then leave the room for half an hour. If they had an answer by the time he returned, he would accept them.It's too long since I did any general relativity to recall all the details, but ...

<t>I still say it can't be done: they have to guess their hat colour simultaneously, and thus cannot use any information from other prisoners' guesses when making their own. Are you sure you don't mean sequentially?As the question stands, for only a finite number of incorrect guesses, each prisoner ...

<t>Is the line infinite in both directions? ie when a prisoner looks at the hats in front of him, can he see a finite or infinte number?Not sure the answer helps anyway. If the prisoners have to guess the colour of their hats simultaneously, all they can go on is the colour of the hats in front of t...

<t>I think changing the BC at an isolated point by a finite amount will not change the solution to the equation. Thinking about things physically, you are not adding any extra 'mass' to the initial condition by changing the 'density' at a point.If I remember rightly there is some theorem in analysis...