# Tag Info

11

It all depends on your level of risk aversion and degree of intertemporal substitution. Let's assume you are risk neutral: Game is played once: you are willing to pay $6.5 = \sum^{N=12}_{i=1} \frac{1}{12} i$ Game is played 10,000 times. Still willing to play 6.5$for each game. Card replaced: well you replace everytime your first draw is lower than 6.5. ... 7 Really you need a degree Reading any one book from the above will not set you up. Furthermore, you will find yourself trapped in a cycle, where really none of the books you suggested can be read in isolation. Taking for example a book on PDEs, you will quickly find you need a lot of knowledge of linear algebra if you want to approximate any of these. For ... 3 I think you have it backwards regarding how much to bet if you play once vs. many times. The optimal amount to bet is given by the Kelly criterion as you said. But you should be MORE inclined to bet closer to the optimal Kelly fraction if you get to play many times. The more you play the more your outcome will approach the expected outcome (which is where ... 3 If the game is played in exactly the way you stated it, why would you ever bet more than 1 dollar? Assuming you bet 1\$, then you get 1\$x value on card. And if you bet 12\$, you get 1\x Value on card. What's the point of betting more than 1 dollar? 3 Note that \begin{align*} \frac{S_T-S_t}{S_t} &= \frac{S_T-K +K-S_t}{S_t}\\ &=\frac{(S_T-K)^+-(K-S_T)^+ +K-S_t}{S_t}. \end{align*} Then, \begin{align*} E\left(\frac{S_T-S_t}{S_t} \mid \mathcal{F}_t \right) &= \frac{e^{rT}}{S_t}(C_t-P_t)+ \frac{K-S_t}{S_t}. \end{align*} where \begin{align*} C_t &= e^{-rT} E\left((S_T-K)^+ \mid \mathcal{F}_t \... 3 In general, the variance of a portfolio is just $$\sigma_p^2 = \sum_i \sum_j w_i w_j \sigma_i \sigma_j \rho_{ij},$$ which intuitively makes sense since we are summing over all weighted standard deviations and their correlations. Sincew_i = \frac{1}{3}$and$\sigma_i = \sigma$for all$i = \{1,2,3\}$, and$\rho_{ij} = 0$for all$i \neq j$, it simplifies to ... 3 The optimal investment strategy depends on the investment goals, or equivalently your utility function (which the investment strategy is supposed to maximize). The forward will trade at$\mathbf{E}^*_0(F_T)$in the market when you invest at$t=0$. If you buy your maximum volume$M$, then gain/loss at$T$is given by$M(F_T-\mathbf{E}^*_0(F_T))$(which is ... 3 Note that \begin{equation} E\big[e^{\sigma \alpha \sqrt{T} N(0,1)}\big] = e^{\frac{\sigma^2 \alpha^2}{2}T} \end{equation} Hence$F_T(T)^\alpha$will be a lognormal variable with expected value$F_T(0)^\alpha e^{-\frac{1}{2}\sigma^2T \alpha + \frac{1}{2}\sigma^2 \alpha^2T}$and log-variance$\sigma^2 \alpha^2 T$. Compare this to the Black formula for ... 3 The idea is pretty much the same as the one used in the Breeden-Litzenberger result. You'll find many questions related to this here already, see e.g.: Prove that the butterfly condition is always greater than zero. The current value of the derivative is the discounted expected value. \begin{equation} D_0 = e^{-r T} \int_K^\infty (x - K)^2 f(x) \mathrm{d}x,... 3 a.) The market capitalization$m_{cap} = 100*\$1.50 + 150*\$2.0 = \$150 + \$300 = \$450$, so the weight of each asset is $1/3$ and $2/3$ respectively in the market portfolio. You don't need to find the minimum variance portfolio. If you plug in these values you get exactly $E[r_m] = 1/3*0.15 + 2/3*0.12 = 0.13.$ b.) The formula is wrong, as you multiply ...

2

The standard formula for Capital Asset Pricing Model is: \begin{equation} \bar{r} = r_f + \beta \cdot ( \bar{r_m} - r_f) \quad (1) \end{equation} in which: \begin{equation} \bar{r} \textit{ - expected return of an asset} \end{equation} \begin{equation} \ r_f \textit{ - risk-free rate} \end{equation} \begin{equation} \beta \textit{ - beta of an asset} \end{...

2

Find the conditions under which: $E_{0}^{*}[\max (P_{T} - HR\times G_T, 0)] = \max (P_{0} - HR\times G_0, 0)$ We have a no-brainer solution - the condition that the drift and volatility of both $P$ and $G$ is zero, which means $P$ and $G$ are constants in time. Second valid condition - the option is deep in the money or deep out of the money, such that ...

2

\begin{equation} \mathbb{E} \left[ \left. W_s \left( W_t - W_s \right) \right| \mathfrak{F}_s \right] = W_s \mathbb{E} \left[ W_t - W_s \right] = 0 \end{equation} The first step uses that $W_s$ is $\mathfrak{F}_s$-measureable and that the increment $W_t - W_s$ is independent of $\mathfrak{F}_s$. Next, \begin{equation} \mathbb{E} \left[ \left. \left( W_t - ...

2

Because instantaneous variance can be written as follows: $V \left[ dS_t\right]=E\left[ \left( dS_t -E\left[dS_t\right] \right)^2\right]$ $V \left[ dS_t\right]=E\left[ \left( dS_t -f \, dt \right)^2\right]$ $V \left[ dS_t\right]=E\left[ \left( g \, dW_t \right)^2\right]=g^2dt$ Which is the same thing as: \$V \left[ dS_t\right]=E\left[ dS_t dS_t\right]=g^...

2

This is pretty standard fare for a Stats 101 course, so as to rationale, etc. you might benefit from picking up a textbook or otherwise do some reading on this. In brief though, hypothesis testing allows us to assess the likelihood sample estimates are different than theorized values in the absence of actual population values. In the cases above, with a ...

2

No, this is in general not true. For example, consider \begin{align*} y(x)&:=x,\\ x(a,b):&=a+b. \end{align*} Then we have $$\frac{\partial y}{\partial x}=1$$ but $$\frac{\partial y}{\partial a} + \frac{\partial y}{\partial b}=2.$$

2

To elaborate on my comment: with respect to questions 1 and 2, the distribution of the payoff for one game is discrete uniform with a mean of 6.5 and a SD of about 3.5 according to my calculations. Now, if you are guaranteed to play this 10,000 times, then you are enititled to consider the distribution of the sum of the outcomes, which by the Central Limit ...

2

In the third equation they somehow did not put the 1/3 in front of the second ln-term. Without the 1/3, the equation looks at follows: error_term_3 = 3 * (np.log(atm_strike / forward) ** 2) * (np.log(atm_strike / forward) - 1 + (forward / atm_strike)) Filling in F0 = 1106.85 and K0 = 1105 gives Epsilon3 = 1.1752170209137409e-11, which is the answer you ...

1

For continuous play you generally want to maximize the expected log return (time average rate of growth under geometric wealth dynamics) conditional on not going bust. This is basically Kelly setup without the approximation. Assume you can bet fractional amounts to simplify: In : import sympy as s In : J = s.log(1 + a * (3 - 1)) * 0.5 + s.log(1 + ...

1

Off the top of my head, if interest rates are zero or negative, then yes. Just borrow the purchase price and buy any asset. Sell later and pay off the loan. Otherwise no.

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