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Let's say that we have a probabilistic forecast for the future percentage return of an asset in the form of a probability density, $\hat{R}_{t+1}$.

If our initial goal was to create a probabilistic forecast for the future price of this asset in the form of a probability density, $\hat{p}_{t+1}$, but we resorted to probabilistically forecasting the future percentage return, $\hat{R}_{t+1}$, since this is usually less involved, is there a way we can backtransform $\hat{R}_{t+1}$ to $\hat{p}_{t+1}$ if the relation between percentage returns and prices is as follows:

$$R_{t+1} = \frac{p_{t + 1}}{p_t} - 1$$

and we know $p_t$? Especially without assumptions on our density functions?

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    $\begingroup$ If it is true that $\hat{p}_{t+1}=\hat{p}_t\cdot\left(\hat{R}_{t+1}+1\right)$, and that such $\hat{p}_{t+1}$ is an appropriate estimator for $p_{t+1}$, can you not simply sample from the known $\hat{p}_t$, and $\hat{R}_{t+1}$, and combine the samples to get a sample from the distribution of $\hat{p}_{t+1}$? This should be enough for all calculations, and it would not place any assumptions on distribution $\endgroup$
    – Cryo
    Commented Apr 8 at 23:21
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    $\begingroup$ @Cryo thank you, yes, I think that would work, I was just wondering if there was a way to do it specifically using the densities. While looking, I came aross the following for the pdf of a transformed variable given the untransformed density and the transformation: Given that $g$ is either an increasing or decreasing function on the image of $X$ and $Y=g(X)$, $$f_{Y}(y) = f_{X}(g^{-1}(y)) = |\frac{d}{dy}g^{-1}(y)|$$ $\endgroup$
    – QMath
    Commented Apr 9 at 3:35
  • $\begingroup$ I tried this for the case of simple returns in my question, and it seemed to work out (I got $f_{P_{t+1}}(p_{t+1}) = f_{R_{t+1}}(\frac{p_{t+1}}{p_t} - 1) |{p_t}^{-1}|$) $\endgroup$
    – QMath
    Commented Apr 9 at 3:40
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    $\begingroup$ However, when I tried for the case of log returns $(r_{t+1} = \text{ln}(\frac{p_{t+1}}{p_t}))$, I got the unintuitive (to me) answer of $f_{P_{t+1}}(p_{t+1}) = f_{R_{t+1}}(\text{ln}(\frac{p_{t+1}}{p_t}))|{p_{t+1}}^{-1}|$ since I feel like the exponential function should come in somewhere? $\endgroup$
    – QMath
    Commented Apr 9 at 3:49
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    $\begingroup$ You said initially you did not want assumptions on density functions, but now you seem to want to derive distribution of $x=y\cdot \left(r+1\right)$, given distributions of $r$ and $y$ (I have changed the labels for convenience). The latter is a 2d problem. So a good way to treat it is to introduce another variable, say $z=\left(r+1\right)/y$, consider coordinate change $\left(y,r\right)\to\left(x,z\right)$, using appropriate Jacobian, then marginalize out the $z$ to get marginal distribution for $x$. Basically, your full derivatives should be partial derivatives in this case $\endgroup$
    – Cryo
    Commented Apr 9 at 10:58

1 Answer 1

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As stated in one of your comments, this is straightforward in the case of an invertible and differentiable transformation $g$, i.e. if we know the density of $X$ and $Y = g(X)$. Then the density of $Y$ is given by $$ f_Y (y) = f_X(g^{-1} (y) ) \frac{d g^{-1} (x)}{dx} = \frac{f_X(g^{-1} (y) )}{g' (g^{-1} (y))} $$ (note the typo in your comment).

In your specific case, if we know the price of today $p_t$ then

  • $g(x) = p_t (x + 1)$ for percentage returns. The resulting density of $p_{t+1}$ is $$ f_{p_{t+1}} (y) = \frac{1}{p_t} f_{R_{t+1}} \left( \frac{y}{p_t} - 1 \right) $$

  • $g(x) = p_t \exp(x)$ for logarithmic returns (that's where the exponential function is you are looking for.) As density I get $$ f_{p_{t+1}} (y) = \frac{1}{y} f_{R_{t+1}} \left( \log \left( \frac{y}{p_t} \right) \right) $$ since $g = g'$ in this situation.

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  • $\begingroup$ Great, thank you for the clarification, are you able to check if my answers I got to in the comments of my original post were the correct ones? $\endgroup$
    – QMath
    Commented Apr 10 at 1:35
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    $\begingroup$ @QMath Yes, I got the same densities, see my edits. $\endgroup$
    – picky_porpoise
    Commented Apr 10 at 19:19
  • $\begingroup$ Great, thank you so much for adding! :) $\endgroup$
    – QMath
    Commented Apr 11 at 2:57

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