8

$\max(B_T,S_T)=\max(0,S_T-B_T)+B_T,$ so this is just a call option (with strike $B_T$) plus $B_T.$


6

No reply has been given so I wanted to at least give a visualisation of the expectiles. Suppose the curvy dashed line in my picture represents a cumulative distribution function of some random variable X. Then blue part corresponds exactly to $\mathbb{E}[(X-x)_+]$, while the orange surface corresponds to $\mathbb{E}[(X-x)_-]$. In the picture $x=1$. Now if ...


4

If you make the change of variable $Y_t = \sinh U_t$ and apply Ito then you immediately get $$dU_t = 2dW_t$$ so the solution of your SDE is $$Y_t = \sinh\left(2W_t + C\right)$$ with $C$ a constant. Then to answer your question is suffices to notice that $$\frac{Y_u}{\sqrt{1+Y_u^2}}=\tanh(U_t)$$ which is bounded therefore your expression is finite ...


4

One of the best ways I came across to estimate the expected return of a stock (even with limited time-series data), is Martin and Wagner (2019): What is the expected return on a stock?. From their paper: Second, our formula provides conditional forecasts at the level of the individual stock. Rather than asking, say, what the unconditional average ...


3

On average half the time you will win 0.5 times your current bankroll and half the time you will lose 0.5 times your current bankroll. Over N plays your expected growth will be (0.5)^(N/2)(1.5)^(N/2) and you will tend to lose money over time and in the limit since 0.5*1.5 = 0.75 < 1. This happens because gaining and losing 50% are not equivalent. Think ...


3

Expectations and conditional expectations are either random or fixed points. It depends upon your choice of axioms. Unfortunately, I have never found a good book that fairly describes both well. On the Bayesian side, I would suggest the very polemical "Probability Theory: The Language of Science" by E.T. Jaynes. On the null hypothesis side, I would ...


3

The expectation looks correct, assuming the function in front of the Brownian is deterministic. It is a standard result in stochastic calculus that the expected value of the integral of a deterministic function with respect to the Brownian motion is zero. You may want to check the properties of the stochastic integral, one of which is the property that I ...


3

The integral $I_T$ is an Itô stochastic integral therefore its expectation is $0$. This is because $I_T$ is a martingale (see e.g. Theorem 4.3.1 in Shreve), hence: $$\mathbb{E}[I_T]=I_0=0$$ You can also see this by considering the definition of a stochastic integral, which involves the sum of terms of the form $f(W_{t_i})(W_{t_{i+1}}-W_{t_i})$, and using the ...


3

This holds due to a change of measure. There is the real-world $\mathbb{P}$ and the risk-neutral world $\mathbb{Q}$. (I am going to assume constant interest rate $r$) The first fundamental theorem of asset pricing states that if there are no arbitrage strategies in a market, then there exists at least one probability measure $\mathbb{Q}\sim\mathbb{P}$ such ...


3

So i'm kinda guessing what you really mean by the logarithmic mean - i'm guessing you mean the logarithmic average of returns - where you mean geometric average. $$ \left( \prod_{i=0}^n a_i \right)^{\frac{1}{n}} $$ where $a_i$ are our returns. We have to make an assumption here - that your underlying is described by $\mathrm{d}S = \mu S \mathrm{d}t + \...


2

$I(t)=\int_0^t \sqrt tdW_s=\sqrt t \int_0^t dW_s =\sqrt t W_t $ and then $$E(I(t)^4)=E(t^2 W_t^4)=t^2 \cdot 3t^2=3t^4$$ using the 4th moment of the $N(0,\sigma^2=t)$ distribution.


2

If you write the SDE in the integral form everything should be straightforward: $$ X_t = X_0 + \int_0^t a(X_s, s) ds + \int_0^t b(X_s, s) dz_s $$ If you now take the expected value the third term disappear since $A$ has $0$ mean (and by definition of the integral of stochastic process). At the end you obtain: $$ \mathbb{E} \left[ X_t\right] = X_0 + \...


2

Here is a list of books I have and like: Pretty much every puzzle style question I got in any interviews I'd seen before, essentially from reading books like these. puzzlegrams, pentagram Simpler puzzles, nicely illustrated. My Best Mathematical and Logic Puzzles (Dover Recreational Math), Marten Gardner A nice book of shortish puzzles, harder than ...


2

Let $$Z_t=Y_t\int_0^t\frac{a}{Y_s}ds$$ Then $Z_0=0$. We differentiate $Z_t$ and obtain $$dZ_t=\int_0^t\frac{a}{Y_s}dsdY_t+Y_t\frac{a}{Y_t}dt=\int_0^t\frac{a}{Y_s}ds(rY_tdt+\sigma Y_td\tilde{W_t})+adt$$ $$=rY_t\int_0^t\frac{a}{Y_s}dsdt+\sigma Y_t\int_0^t\frac{a}{Y_s}dsd\tilde{W_t}+adt$$ Then $$dZ_t=rZ_tdt+\sigma Z_td\tilde{W_t}+adt$$ We have $$d(e^{-rt}Z_t)=...


1

You state $X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$. Thus: \begin{equation} X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z , \end{equation} where $Z \backsim \mathcal{N}(0, 1)$. Note that: \begin{align} & \mathbb{E} \sqrt{\frac{t}{n}} Z = 0 \\ & \mathbb{E} \left( \sqrt{\frac{t}{n}} Z \right)^2 = \frac{t}{n} \mathbb{E} Z^2 = \frac{t}{n} \\...


1

Your process for $(S_t)$ is a geoemtric Brownian motion and since $S_t=S_0 e^{\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t}$, we have \begin{align*} \ln(S_t) &= \ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t \\ &\sim N\left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t,\sigma^2 t\right). \end{align*} Thus, \begin{align*} X_t &= e^{\...


1

No because $$ E(e^Z)=e^{\frac{1}{2}}\neq1 $$ More generally: $$ N \sim \mathcal N(\mu,\sigma^2)\\ E(e^{Nt})=MGF_{\mathcal N(\mu, \sigma^2)}(t)=e^{\mu t+\frac{1}{2}\sigma^2t^2} $$ The last lines should be: $$ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =\\ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})e^{\frac{1}{2}} $$ For the rest it is correct.


1

Well, in a standard contract, if you bumped the curve, this will affect the swap value (which can be negative). However, since the swaption payoff is $Max(V_{swap},0)$ , where $V_{swap}$ is the swap value, this cannot be negative.


1

Note that because the integrand is deterministic this Itô integral is normally distributed with parameters (cf. Itô isometry) $$ I_t := \int_0^t \sqrt{s} dW_s \sim N(0, t^2/2) $$ Now you can just use the results that apply for the moments of a Gaussian variable.


1

Let $(W_t)_{t \geq 0}$ denote a standard Brownian motion and $\Bbb{F}=\{\mathcal{F}_t\}_{t \geq 0}$ the natural filtration it generates over some probability space $(\Omega, \Bbb{P})$. By definition, we know that $$\forall 0 < s < t, W_t - W_s \sim \sqrt{t-s}\, N(0, 1)$$ under $\Bbb{P}$. Noting that $W_t = (W_t - W_s) + W_s$ and conditioning on the ...


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