11 votes
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Expectation of $\int_0^t \frac{1}{1+W_s^2} \text dW_s$

By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$. The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{...
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  • 13.8k
10 votes

How to compute $E[W(T)\exp(W(T)]$

Hereunder is how I would solve that. I would say this is some sort recurring exercice in probability classes at university. Solution based on the derivation of the characteristic function $e^{\lambda ...
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8 votes

How to compute $E[W(T)\exp(W(T)]$

Notice that $$e^{W^{Q}(T)}$$ looks almost like Doleans exponential $$e^{W^{Q}(T)-\frac{1}{2}T}$$ Therefore $$E^Q[W^{Q}(T)e^{W^{Q}(T)}]=E^Q[W^{Q}(T)e^{W^{Q}(T)}]e^{-\frac{1}{2}T}e^{+\frac{1}{2}T}$$ $$=...
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  • 816
7 votes
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Intuitive explanation for expectiles

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 ...
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  • 1,467
6 votes
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How to check if $ E [\exp \{ \int_0^t \frac{Y_u^2}{1+Y_u^2}du \}]< \infty $

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. ...
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  • 2,117
6 votes

How to compute $E[W(T)\exp(W(T)]$

To continue your thought: $$f(x)=x{\rm e}^x $$ $$df(W_t) = \left({\rm e}^{W_t} + f(W_t)\right) dW_t + \left({\rm e}^{W_t} + 1/2f(W_t) \right)dt $$ We now integrate from $0$ to $T$ ($W_0=0$): $$f(W_T) ...
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  • 5,028
6 votes

How to compute $E[W(T)\exp(W(T)]$

So let‘s add the brute force solution as well: $W_T\sim N(0,T)$ so $$ \begin{align} E\left(W_Te^{W_T}\right)&=\int_{-\infty}^{\infty}xe^x\frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi T}}dx\\ &= \int_{-...
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  • 5,763
4 votes

How to calculate the expected stock returns for an individual stock?

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 ...
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  • 6,860
4 votes
Accepted

Show that $\mathbb{E}[(S+\xi)^2]\rightarrow 0$ as $n\rightarrow\infty$

Note that \begin{align*} \int_0^T W(t)dt \approx \sum_{j=0}^{n-1}\frac{T}{n}W\Big(\frac{jT}{n}\Big). \end{align*} Then, \begin{align*} &\ \sum_{j=0}^{n-1}\frac{jT}{n}\bigg(W\Big(\frac{(j+1)T}{n}\...
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  • 20.4k
4 votes

How to compute $E[W(T)\exp(W(T)]$

There are many roads to Rome. Here is the road of overkill, but which nevertheless gives a glimpse of the uses of Malliavin calculus: Note that $$ E_0 \left[W(T) e^{W(T)} \right] = E_0 \left[e^{W(T)} ...
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3 votes

Question about slides in lecture note: What if we can't assume $\mu=0?$

I feel that question 1 has already been answered in the comments, so I will provide a brief answer and insight for question 2 (which is also your title question). Question 2: What happens when $W \sim ...
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  • 3,618
3 votes

What's the expected value of a repeated game with 50% chance to win 0.5 and 50% to lose 0.5?

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) ...
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  • 909
3 votes
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How to calculate the mean and variance of this Ito integral?

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 ...
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3 votes
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Proof that we can price any derivative as the discounted value of its expected return under the risk neutral measure

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 ...
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  • 13.8k
3 votes
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Condition expectation calculation examples and theory

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. ...
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  • 4,092
3 votes

Expected value of stochastic optimization

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 ...
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3 votes
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Why is logarithmic mean equal to the arithmetic expectation less one-half its variance?

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)^{\...
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  • 2,406
2 votes

Fourth moment of a itos integral

$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.
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2 votes

Expected Value of Stochastic Process

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 ...
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  • 445
2 votes

Reference material (EV/ betting game questions) for Quant Hedge Funds Interviews

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 ...
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  • 2,406
2 votes
Accepted

Log-normal risk-neutral price derivation from binomial trees, not clear about step in derivation process

Let $X\sim N(0,1)$ be a standard normal variable and $\alpha:=\sigma\sqrt{T}$, then by definition of the expectation and the distribution of normal variables: $$\begin{align} \mathbb{E}\left(e^{\alpha ...
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2 votes

Properties of integrated GBM

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$...
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2 votes

Question about slides in lecture note: What if we can't assume $\mu=0?$

(Q2) Adding a constant $\mu$ to the white noise, we have $$Y_t := (\mu+W_t) + b(\mu + W_{t-1}) = \mu(1+b) + X_t, $$ which is MA(1) with drift: $$E[Y_t] = \mu(1+b) $$ and $$ {\rm Cov}(Y_t, Y_{t-k}) = {...
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  • 5,028
2 votes
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Esscher Premium: Integral Transform Proof

Okay, I've tried to come up with a solution. We know that $\int x \: dF(x)$ is a generalization of $\int x f(x) \: dx$, since: $$\frac{dF(x)}{dx}=f(x) \iff dF(x)=f(x)\: dx$$ for $f$ being the pdf and $...
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  • 3,618
2 votes
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How to take the expectation of an exponential martingale? And an exponential with a random value?

Given that $exp(\sigma m-\frac{1}{2}\sigma^2 \tau_m)$ is a martingale, you just need to substitute $m = 0$ into it to find the value of its expectation, as for any martingale $Z_t$ we have that $E[Z_t]...
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  • 78
2 votes
Accepted

Show that $\text{Cov}[Z_t,Z_{t+h}]=\text{Cov}[Z_s,Z_{s+h}].$

Note that: $$ Z_tZ_{t+h} = \left(\sum_{i=0}^{n}a_iX_{t-i}\right) \left(\sum_{j=0}^{n}a_jX_{t+h-j}\right) \not =\sum_{i=0}^{n}a_i^2X_{t-i}X_{t+h-i} $$ Yes, as the expectation operator is linear, all we ...
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  • 5,028
1 vote
Accepted

What is the expectation of a change in Brownian motion?

$E(X)=\mu$ doesn't necessarily imply $E[f(X)]=f(\mu)$. In this case, if $X \sim N(\mu,\sigma^2)$ then $e^X \sim lnN(\mu,\sigma^2)$ (lognormal distribution) and $$E(e^X)=e^{\mu+\frac{\sigma^2}{2}}$$. ...
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1 vote

How to Evaluate Expected Value powered 4 of a Wiener Process?

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:...
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1 vote

change of measure expectation

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}\...
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  • 13.8k

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