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In the following image :

enter image description here

  1. I am not able to understand how, the final value of strategy B can be equal to $e^{r_{GBP}T}F(0,T)$

  2. According to me it should be just $F(0,T)$

  3. My reasoning is that when you received 1 pound and invested it, you’ll get $e^{r_{GBP}T}$ pounds in return. But we use that to cover our short position in the forward contract entered at time 0, so at the end we’re only left with the forward price $F(0,T)$ paid to us at time T.

Please correct me where I am going wrong.

Source of image : Mathematical Finance : An Introduction to Financial Engineering by Marek Capinski and Tomasz Zastawniak

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  • $\begingroup$ Got got it. I didn’t know that by F(0,T) we mean the forward contract on 1GBP, I thought F(0,T) in this context means forward contract on $e^{r_{GBP}T}$ GBP. But now it’s clear, thanks alot. If you want, you can write the same as an answer, that’ll allow me to accept it. $\endgroup$ – Vishaal Sudarsan Apr 30 '20 at 10:42
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What is slightly confusing here is where they mention that you should short $X$ forward contracts. Implicitly they are assuming that each "contract" refers to one GBP and you buy or sell as many contracts as you need. (In practice you would simply call the dealer and tell them the size of the GBP position you have in mind, no need to mention "contracts").

Now the analysis. The 1 Pound mentioned in B. will have grown to $1 \cdot e^{r_{GBP}T}$ Pounds at time T. So you must today sell forward not 1 forward contract but $e^{r_{GBP}T}$ of them, so you are able to convert at time T the full amount of GBP that you have and not just the initial 1 GBP, into USD. This justifies the RHS of the equation shown.

Another way to look at it (without mentioning contracts) is that if $F(0,T)$ is the forward rate then $X \cdot F(0,T)$ is the result of converting $X$ GBP into USD at this rate. Here $X$ is $1 \cdot e^{r_{GBP}T}$.

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