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The Laplace transform of the integrated process CIR process is given by, see e.g. Dufresne (2001). you can download it The integrated square-root process

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As Sherev has said, first let $\varphi(t)=E\left[e^{tX}\right]$ then $$\varphi '(t)=\underset{s\to t}{\mathop{\lim }}\,\frac{\varphi (t)-\varphi (s)}{t-s}=\underset{s\to t}{\mathop{\lim }}\,\frac{E[{{e}^{tX}}]-E[{{e}^{sX}}]}{t-s}=\underset{s\to t}{\mathop{\lim }}\,E\left[ \frac{{{e}^{tX}}-{{e}^{sX}}}{t-s} \right]$$ Sherev continue ,we can choose a sequence ...

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That formula actually does not make much sense, given that for a continuous random variable the probability of any given point is zero. Assuming a Black-Scholes world you are better of by doing: $P(S_T>S_T^*)=N(d_2)$ where $d_2$ is the standard black-scholes term. From this it is straightforward to compute $P(S_T^{-}<S_T<S_T^+)$.

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Perhaps other memeber of qSE are going to correct me, but I think the following rule of thumb is useful. Whenever you have a doubt, try to forget that a pricing measure is a probability measure. This is just a pricing tool: originally for any option/derivative/contingent claim we'd like to know its price, so we introduce a map $\pi:X\to \Bbb R$ such that ...

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This is a standard way of defining a product measure in measure theory. The reasoning is a bit technical, but pretty natural. Let's say you have two measure spaces $(X,\mathcal A, \mu)$ and $(Y,\mathcal B, \nu)$, and you would like to construct a product measure space $(Z,\mathcal C, \phi)$ where $Z = X\times Y$, and $\mathcal C = \mathcal A\otimes\mathcal ... 1 For the terminal distributions, I don't have the closed-form solution to hand, but it's computable, since we can price power options (with payoffs like$(S_T^n-K)^+$). You need to find $$E[S_T C_{K,T}] = \int_K^\infty x(x-K) \cdot p_{BS}(x) dx \\=-Ke^{(r-q)T} C_{K,T} + \int_K^\infty x^2 \cdot p_{BS}(x) dx$$ The latter formula is just a power-option ... 0 Here is a sketch for an argument for 'yes': Let$\Omega$have$n$elements. For each extra element we add, the smallest$\sigma$-algebra containing$\Omega$,$\sigma(\Omega)$, will add some finite number of elements. Now consider the rational numbers. Between each pair of consecutive natural numbers there are infinitely many rational numbers, so each of the ... 3 Your question is not clear. What you might want to say is what distribution should the futures price follow, under the risk-neutral or physical probability measure. In this sense, it will depend on your intention. For potential future exposure, you may want to use the physical measure for the price evolution, while the distribution will depend on your model ... 0 As the Radon-Nykodim derivative is defined for measures on the same measurable space, and while probability$P$is defined on the probability space$\Omega$, the standard machine is necessary to have the two measures$\lambda$and$P \circ X^{-1}$both defined on$\mathbb{R}\$. That is, \begin{align*} \mathbb E(g(X)) &=\lim_{n\rightarrow ...

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