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For most instruments will the EE(t) increase with growing t. If you have very high PDs the probability of defaulting at an early time rises, so that there there is not enough time to develop an exposure. Imagine a swap that is worth zero at inception. For very high hazard rates, a default will...

Hi Brian, you can use bootstrapping with a sample of random vectors or just a sample of random numbers. It doesn't matter. Your row 7 would be a sample of random numbers. E.g. You can calculate the standard derivative of that sample and use Bootstrapping to estimate the quality of that...

You can't get from A to B because there is no way. I don't know who made an error, but if you insert a plus sign into the first formula instead of a multiplication, you can get to B: ( D * e^-rt - Put ) + e^yt = D => e^yt = D - ( D * e^-rt - Put) => e^yt = D * (1 - e^-rt) + Put => y = 1/t * ln(...

- Discrete distribution: There is no pdf. - Use of pdf: The pdf is indeed the derivative of the cdf. You need it for example to calculate the expected value of a continuous distribution, which is E(x) = \int x * f(x) dx with f(x) being the pdf.

Hi cabrown085, sorry, you got some stuff mixed up. A pdf for a discrete distribution is almost everywhere zero (and mostly not very useful). pdfs are used with continuous distributions. The advantage of cdfs is indeed, that they can be used for discrete and continuous distributions. Anyway...

You just seem to have a slight problem with the terms: - for continuous distribution the density is the curve (e.g. The blue curve in my picture. The density does not denote probability directly, but probability density. That means the area under the curve for a given interval is the...

Hi Brian, mass and density are not the same. All the slices have the same mass by definition, which is 0,5% probability, but the probability density in each slice might very. I made this little picture to illustrate, which shows the density of a standard normal distribution. I marked the...

Also to add to my above post: I took a quick look at your graphic of your calculations. I think you made an error in counting your wins and losses. On case of a win you count the stake money as win. But you just got the money back, you placed earlier. Look at your first 3 games and assume you...

This is a variant of the classical martingale strategy: https://en.m.wikipedia.org/wiki/Martingale_(betting_system) The good news is, you will almost surely win with this strategy, the bad news is, you need infinite wealth and infinite number of games for that to happen. With a finite amount of...

I think var(X + Y) = var(X) + var(Y) is only true if X and Y are independent. In general it is var(a*X + b*Y) = a^2 * var(X) + b^2 * var(Y) + 2*a*b*Cov(X,Y) This formula can be derived by writing down the definition of var() and Cov() and using the linearity of the expected value: E(a*X + b*Y)...

You can use the square root rule: vol(20) = vol(2) * sqrt(20/2) = 3.79%

As I said, I'm not sure how determinant is defined in this context, but I doubt that Z is it. The regression coefficient of Z is zero in this model because X1 and X2 already explain 100% of the variability of Y. If Z = X2^3 they are correlated. If Z = X2^2 they are only correlated, if you...

Hi Brian, I'm pretty sure only linear correlation counts here. If the omitted variable is not correlated to the other independent variables, it can be seen as an error term. I seem to recall, that error terms need to be uncorrelated, but not necessarily statistical independent. I don't have any...

As I see it, the problem is, that linear models are not uniquly defined in their independent variables. If your model is: b1 * X1 + b2 * X2 = Y Then also b1 * X1 + b3 * X3 = Y with X3 = 2 * X2 is an equivalent model. I would define the conditions for OVB like this: If we consider the...

The difference is between a loan to a borrower with no risk to default and a risky borrower. Im reality a borrower with no risk doesn't exist, but there are approximations like AAA rated Banks or certain goverments. If a riskless entity has to pay x% on a loan and you have to pay y% than y% -...

It always helps me, to think of the foreign currency as commodity, that is bought or sold with money in the local currency. So if you want to calculate forward rates for USD/BRL, which is the price of 1 USD in BRL, then BRL is the local currency and USD is foreign currency. So the current price...

You are right in the fact, that the price in d1 has to be discounted with the dividends too. This discounting is already included in your formula in the r - q term. For better understanding, rewrite the numerator in d1 as: ln( S0/K * exp( (r - q + σ^2/2) * T) ) you see, the dividend is indeed...

I can't find your formulas in my (old) edition of Hull, and I can't see the study note, but let me try to answer anyway: Formula 1 is the mean geometric return of a geometric Brownian motion (I assume it is μ - σ^2/2). To be precise, it is the mean of the logarithm of a geometric Brownian...

In d1 the S0 in ln(S0/K) is the current price. The meaning is consistent in all formulas that you are citing. What gave you the impression, that S0 is differently defined somewhere? Also compare https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model

The t-distribution with unlimited df is identical to the normal distribution i.e. Z-table. The 1.96 is valid for a two tailed z-test with 5% confidence level. I guess the rationale for using the z-test here is, that the variance is known. Because we actually deal with a binominal distribution...