User qp212223 - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2023-02-01T19:00:45Z https://quant.stackexchange.com/feeds/user/49970 https://creativecommons.org/licenses/by-sa/4.0/rdf https://quant.stackexchange.com/q/59716 2 Risk neutral probability for stock with continuous dividend qp212223 https://quant.stackexchange.com/users/49970 2020-12-02T22:37:42Z 2022-06-08T06:31:08Z <p>Setting: binomial tree with one step over time <span class="math-container">$\Delta t$</span>. I'm trying to derive the risk neutral probability for a stock which pays a continuous dividend, say <span class="math-container">$\delta$</span>. i.e. probability <span class="math-container">$p$</span> such that <span class="math-container">$$e^{r \Delta t} S_0 = S_u p + S_d(1-p)$$</span></p> <p>where <span class="math-container">$S_u, S_d$</span> are the values of the stock in the up and down states respectively. This immediately gives <span class="math-container">$$p = \frac{S_0 e^{r \Delta t} - S_d}{S_u - S_d}$$</span></p> <p>Now if we assume <span class="math-container">$S$</span> has volatility <span class="math-container">$\sigma$</span>, we should be getting <span class="math-container">$S_d = S_0 e^{-\sigma \sqrt{\Delta t} - \delta \Delta t}$</span> and <span class="math-container">$S_u = S_0 e^{\sigma \sqrt{\Delta t} - \delta \Delta t}$</span> so that <span class="math-container">$$p = \frac{e^{r \Delta t} - e^{- \sigma \sqrt{\Delta t} - \delta \Delta t} }{ e^{ \sigma \sqrt{\Delta t} - \delta \Delta t} - e^{- \sigma \sqrt{\Delta t} - \delta \Delta t}} = \frac{e^{(r+ \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} - e^{- \sigma \sqrt{\Delta t}}}$$</span></p> <p>but this is wrong because the formula that's given in my course's lecture notes on this is <span class="math-container">$$p = \frac{e^{(r- \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} - e^{- \sigma \sqrt{\Delta t}}}$$</span></p> <p>(the only difference is the <span class="math-container">$r-\delta$</span> in the numerator instead of the <span class="math-container">$r+ \delta$</span>). I don't understand why my assumptions on the values for <span class="math-container">$S_u$</span> and <span class="math-container">$S_d$</span> are wrong. Any help would be massively appreciated.</p> <hr /> <p>MY POTENTIAL EXPLANATION: perhaps the value of <span class="math-container">$S_u$</span> should be <span class="math-container">$S_0 e^{\sigma \sqrt{\Delta t} + \delta \Delta t}$</span> (and similarly with <span class="math-container">$S_d$</span>) because we work with the <em>payoff</em> of owning one unit of the stock, so if we increase with upward factor <span class="math-container">$e^{\sigma \sqrt{\Delta t}}$</span> we GAIN the value of the dividend, not lose it.</p> https://quant.stackexchange.com/q/60100 1 Delta hedging for an American call option on a stock with a continuous dividend yield qp212223 https://quant.stackexchange.com/users/49970 2020-12-21T20:15:20Z 2021-09-18T10:04:50Z <p>Let the dividend yield be <span class="math-container">$\delta$</span> and <span class="math-container">$C_u, C_d$</span> and <span class="math-container">$S_u, S_d$</span> be the up and down values for the stock and the call respectively over the period <span class="math-container">$\Delta t$</span>.</p> <p>In Hull and all other resources I've looked at, the hedge ratio stays the same in this case as the no dividend yield case, i.e. <span class="math-container">$$\Delta = \frac{C_u - C_d}{S_u - S_d}$$</span> which confuses me because the payoff of owning one share of stock is actually <span class="math-container">$S_u e^{\delta \Delta t}$</span> or <span class="math-container">$S_d e^{\delta \Delta t}$</span> so I would presume that the hedge ratio should shift to be <span class="math-container">$$\frac{C_u - C_d}{S_u - S_d} \exp (-\delta \Delta t)$$</span> Why is this not the case?</p> <hr /> <p><em><strong>My attempt at a &quot;reasonable&quot; explanation:</strong></em></p> <p>Over short <span class="math-container">$\Delta t$</span> we have <span class="math-container">$\exp(\delta \Delta t) \approx (1+ \delta \Delta t)$</span> so that the approximate payoffs from the stock are <span class="math-container">$S_u + S \delta \Delta t$</span> in the up position and <span class="math-container">$S_d + S \delta \Delta t$</span> in the down position, where <span class="math-container">$S$</span> is the initial price of the stock, asymptotically, so we may just take the denominator in the hedge ratio to be <span class="math-container">$$S_u + S \delta \Delta t - (S_d + S \delta \Delta t) = S_u - S_d$$</span> as usual.</p> https://quant.stackexchange.com/q/59088 3 Application of Ito's lemma relating to bond price qp212223 https://quant.stackexchange.com/users/49970 2020-11-02T05:22:42Z 2020-11-02T20:04:45Z <p>I'm interested in solving the following questions but I am confused on the second part because I do not know how to define/calculate the interest per &quot;unit time&quot;, which I'm guessing is informal terminology for &quot;dt&quot;. I've included my attempt after the picture.</p> <p><a href="https://i.stack.imgur.com/hztRp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hztRp.png" alt="enter image description here" /></a></p> <p>(i) - this is trivial: <span class="math-container">$B(y) = \int_0^\infty e^{-yt} dt = y^{-1} e^{-yt}|_{t=0}^{t = \infty} = \frac{1}{y}$</span></p> <p>(ii) - I'm not sure how to define the interest. Would this be the interest payment <em><strong>received</strong></em> from the bond payment? i.e. informally <span class="math-container">$\exp(-y_t t)dt$</span>?</p> <p>In this case we get <span class="math-container">$$\text{Exp. total return per unit time} = \frac{dB_t + e^{-y_t t}dt}{B_t}$$</span></p> <p>and from Ito's lemma we obtain <span class="math-container">$$dB_t = \frac{-1}{y_t^2} dy_t + \frac{1}{2} \frac{2 d\langle y \rangle_t}{y_t^3} = -\frac{a(m-y_t) dt + by_t dZ_t}{y_t^2} + \frac{b^2}{y_t} dt = \frac{\bigg((b^2+a)y_t - am\bigg) dt + by_t dZ_t}{y_t^2}$$</span></p> <p>so that</p> <p><span class="math-container">$$\text{Exp. total return per unit time} = \frac{\bigg(e^{-y_t t}y_t^2 + (b^2+a)y_t - am\bigg) dt - by_t dZ_t}{y_t}$$</span></p> <p>Would this be the right idea? Thanks for any advice!</p> https://quant.stackexchange.com/q/59036 1 Relationship between risk and return for GBM and riskless bond qp212223 https://quant.stackexchange.com/users/49970 2020-10-30T18:15:03Z 2020-10-30T18:15:03Z <p>Suppose we have <span class="math-container">$S$</span>, a stock following geometric Brownian motion (<span class="math-container">$dS_t = S_t (\mu dt + \sigma dZ_t)$</span> for <span class="math-container">$Z =$</span> Brownian motion) and <span class="math-container">$B$</span>, a zero coupon bond with rate <span class="math-container">$r$</span>, i.e. <span class="math-container">$dB_t = rB_t dt$</span>.</p> <p>In trying to explain/derive the Sharpe ratio using these two assets (<span class="math-container">$= (\mu - r)/\sigma$</span>), a set of lecture notes that I'm reading states that if we invest some proportion <span class="math-container">$w \in [0,1]$</span> in <span class="math-container">$S$</span>, then the expected return is <span class="math-container">$w\mu + (1-w) r$</span> and the volatility is <span class="math-container">$w \sigma$</span> and hence any security with this volatility should give the same expected return. i.e. Any asset with volatility <span class="math-container">$w \sigma$</span> must give return excess of <span class="math-container">$r$</span> of <span class="math-container">$w(\mu - r)$</span> and thus</p> <p><span class="math-container">$$\frac{\text{Excess return}}{\text{Volatility}} = \frac{w(\mu-r)}{w \sigma} = \frac{\mu -r}{\sigma}$$</span></p> <p>This confuses me, because the expected return of the stock is actually <span class="math-container">$\exp(\mu t)$</span> and of the bond is <span class="math-container">$\exp(rt)$</span>. What is the rationale here? I've attached the slide I'm referring to in particular. Is the argument supposed to be purely heuristic over a short period? I've attached the slide I'm interested in below.</p> <p><a href="https://i.stack.imgur.com/ILSXS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ILSXS.png" alt="enter image description here" /></a></p> https://quant.stackexchange.com/questions/59716/risk-neutral-probability-for-stock-with-continuous-dividend?cid=86072 Comment by qp212223 on Risk neutral probability for stock with continuous dividend qp212223 https://quant.stackexchange.com/users/49970 2020-12-03T21:15:33Z 2020-12-03T21:15:33Z This is for the same reason I give in my potential explanation, right? The payoff of owning one share at time $0$ is $S_0\exp(\pm \sigma \sqrt{\Delta t} + \delta \Delta t)$ after $\Delta t$ because of the dividend.