New answers tagged probability
2
As @Martin has pointed out in his answer, of course it is.
Let $X=\sum_{i=1}^N w_ix_i$ denote the return of a portfolio of $N$ assets with multivariate distribution $f(x_1,x_2,\ldots,x_N)$.
The distribution of $X$ may be found by $(N-1)$-fold convolution of the $N$-dimensional distribution $f$. Unfortunately, the integrals are not that easily solved anymore.
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0
As you pointed out, portfolio return is a weighted sum of individual assets returns. Since return of the assets are random variable, naturally portfolio return is also random variable. The distribution of the portfolio return can be infered from distributions of individual assets.
1
Well, I will try to answer 4.
We know that the asset liquidation value $\tilde{v}$ is an affine function of the singals thus we have that
$$\tilde{v}=\bar{v}+\sum_{i=1}^{N}\tilde{s}^i\Rightarrow \tilde{v}=\bar{v}+N\underbrace{\frac{\sum_{i=1}^{N}\tilde{s}^i}{N}}_{\tilde{s}^i}\Rightarrow\tilde{v}=\bar{v}+N\tilde{s}^i$$
where the $\tilde{s}^i$ is the average ...
4
CDS quotes are observable. But none of: probabilities of default, hazard rates, loss given default/recovery, etc are observable.
To get some kind of (risk-neutral) probabilities of default, many people make a lot of assumptions, in particular, that the hazard rate is constant (or if you're lucky enough to have CDS quotes at more than one tenor, then ...
1
Another sketch of proof:
If you move to the equivalent PDE (using Feynman-Kac), you can assume that S is positive, find the solution by log-transfomation.
Then as the solution is unique given initial conditions, and it is the solution of the original PDE, S must be positive.
5
I think the easiest way to derive the solution to the GBM is via Ito's Lemma.
The GBM: $dS_t = \mu S_t dt + \sigma S_t dW_t$ is a short hand for:
$$ S_t = S_0 + \int_{h=0}^{h=t}\left(\mu S_h\right)dh + \int_{h=0}^{h=t}\left(\sigma S_h\right)dW_h $$
Ito process is defined as:
$$ X_t = S_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(...
0
Since a uniform distribution is subgaussian, yes.
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