Log-Concentration Densities and the Jacobian

Why the Jacobian Matters

A concentration $C$ is positive, while its logarithm ranges over the whole real line. That change of scale is useful for describing multiplicative variability and for representing positive distributions with familiar densities. But a density is defined relative to a coordinate: equal intervals in log concentration do not correspond to equal intervals in concentration. The Jacobian accounts for this difference and ensures that probabilities are preserved.

To make the logarithm dimensionless, fix a reference concentration $C_{\mathrm{ref}}>0$, in the same units as $C$, and define $Y=\ln(C/C_{\mathrm{ref}})$. The reference is fixed, not random. Write $p_C(c,t)$ for the density of $C_t$ with respect to concentration $c$, and $p_Y(y,t)$ for the density of $Y_t$ with respect to $y$. Assume initially that $C_t$ is strictly positive and has a density, with no probability mass at a boundary.

Definition: The log-concentration density and concentration density are related by $$ p_Y(y,t)=C_{\mathrm{ref}}e^y\,p_C(C_{\mathrm{ref}}e^y,t), \qquad p_C(c,t)=\frac{1}{c}\,p_Y\!\left(\ln(c/C_{\mathrm{ref}}),t\right),\quad c>0. $$ The factor $C_{\mathrm{ref}}e^y=c$ is the Jacobian $dc/dy$.

These formulas are not optional adjustments to the density. They express the same probability in two coordinate systems. For example, the probability that concentration lies between $c_1$ and $c_2$, where $0<c_1<c_2$, must equal the probability that log concentration lies between $\ln(c_1/C_{\mathrm{ref}})$ and $\ln(c_2/C_{\mathrm{ref}})$.

Result: If $C$ has density $p_C$ on $(0,\infty)$, and $Y=\ln(C/C_{\mathrm{ref}})$, then $$ P(c_1<C\le c_2) =\int_{c_1}^{c_2}p_C(c)\,dc =\int_{\ln(c_1/C_{\mathrm{ref}})}^{\ln(c_2/C_{\mathrm{ref}})}p_Y(y)\,dy. $$ Thus $p_Y(y)=c\,p_C(c)$ at $c=C_{\mathrm{ref}}e^y$.

To verify the result, substitute $c=C_{\mathrm{ref}}e^y$, so $dc/dy=C_{\mathrm{ref}}e^y=c$. The change-of-variables formula gives $$ \int_{\ln(c_1/C_{\mathrm{ref}})}^{\ln(c_2/C_{\mathrm{ref}})} p_Y(y)\,dy = \int_{c_1}^{c_2} p_Y\!\left(\ln(c/C_{\mathrm{ref}})\right)\frac{1}{c}\,dc. $$ For this to equal $\int_{c_1}^{c_2}p_C(c)\,dc$ for every such interval, the densities must satisfy $p_Y(\ln(c/C_{\mathrm{ref}}))/c=p_C(c)$ almost everywhere. Rearranging gives both formulas. In particular, if $p_C$ integrates to one, then $$ \int_{-\infty}^{\infty}p_Y(y)\,dy =\int_0^\infty p_C(c)\,dc=1. $$ The same substitution proves normalization in the reverse direction. $\square$

Worked Density Conversions

Worked Example: A Lognormal Concentration

Let $C_{\mathrm{ref}}=1$ mg/L, and suppose $Y=\ln(C/C_{\mathrm{ref}})\sim N(\ln 2,0.25)$. Find both densities and evaluate them at $y=\ln 2$ and $c=2$ mg/L.

The normal density of $Y$ is $$ p_Y(y)=\frac{1}{0.5\sqrt{2\pi}} \exp\!\left[-\frac{(y-\ln 2)^2}{2(0.25)}\right]. $$ Since $c=C_{\mathrm{ref}}e^y$, the concentration density is $$ p_C(c)=\frac{1}{c\,0.5\sqrt{2\pi}} \exp\!\left[-\frac{(\ln(c/C_{\mathrm{ref}})-\ln 2)^2}{2(0.25)}\right],\qquad c>0. $$ This is a lognormal density. At $y=\ln 2$, $p_Y(\ln 2)=1/(0.5\sqrt{2\pi})\approx0.7979$ per log-concentration unit. At $c=2$ mg/L, $p_C(2)=p_Y(\ln 2)/2\approx0.3989$ per mg/L. The factor $1/2$ is required: the two density values refer to different units of interval width.

Normalization follows directly from the normal density: $\int_0^\infty p_C(c)\,dc=\int_{-\infty}^{\infty}p_Y(y)\,dy=1$. Omitting $1/c$ from $p_C$ would not preserve probability under this substitution.

Worked Example: A Log-Uniform Concentration

Let $Y$ be uniform on $[0,\ln 4]$, with $C_{\mathrm{ref}}=1$ mg/L. Convert its density to concentration coordinates and verify the result.

The log density is $p_Y(y)=1/\ln 4$ for $0\le y\le\ln 4$, and zero elsewhere. The corresponding concentration range is $1\le c\le4$ mg/L. Applying the inverse formula gives $$ p_C(c)=\frac{1}{c\ln 4},\quad 1\le c\le4, \qquad p_C(c)=0\quad\text{otherwise}. $$ Its integral is $$ \int_1^4\frac{1}{c\ln 4}\,dc =\frac{\ln 4-\ln 1}{\ln 4}=1. $$ For example, $P(1<C\le2)=\int_1^2 dc/(c\ln4)=\ln2/\ln4=1/2$. In log coordinates the same probability is the fraction of the interval $[0,\ln4]$ occupied by $(0,\ln2]$, also $1/2$. The concentration density is not constant even though the log density is.

Transforming the Forward Equation

Consider a one-dimensional Itô SDE on a time interval of interest, $$ dC_t=a(C_t,t)\,dt+b(C_t,t)\,dW_t, $$ where $W_t$ is Brownian motion. Assume the coefficients give a non-explosive solution that remains strictly positive on this interval. For the density equations below, also assume the densities and coefficients have enough differentiability for the displayed derivatives to exist, and sufficient decay or boundary behavior for the integrations by parts to have no boundary terms. The standard forward equation for $p_C$ is understood under these conditions.

Theorem: Under these assumptions, $Y_t=\ln(C_t/C_{\mathrm{ref}})$ satisfies $$ dY_t=\alpha(Y_t,t)\,dt+\beta(Y_t,t)\,dW_t, $$ where, with $c=C_{\mathrm{ref}}e^y$, $$ \alpha(y,t)=\frac{a(c,t)}{c}-\frac{b(c,t)^2}{2c^2}, \qquad \beta(y,t)=\frac{b(c,t)}{c}. $$ Its density $p_Y$ satisfies $$ \frac{\partial p_Y}{\partial t} =-\frac{\partial}{\partial y}\bigl(\alpha(y,t)p_Y(y,t)\bigr) +\frac{1}{2}\frac{\partial^2}{\partial y^2} \bigl(\beta(y,t)^2p_Y(y,t)\bigr). $$

Derivation. Itô’s formula applied to $g(c)=\ln(c/C_{\mathrm{ref}})$, with $g'(c)=1/c$ and $g''(c)=-1/c^2$, gives $$ dY_t =\left(\frac{a(C_t,t)}{C_t}-\frac{b(C_t,t)^2}{2C_t^2}\right)dt +\frac{b(C_t,t)}{C_t}\,dW_t. $$ This identifies $\alpha$ and $\beta$. To justify the density equation, take any smooth, compactly supported test function $\varphi$ on the real line. Itô’s formula for $\varphi(Y_t)$, followed by expectation, yields $$ \frac{d}{dt}E[\varphi(Y_t)] =E\!\left[\alpha(Y_t,t)\varphi'(Y_t) +\frac{1}{2}\beta(Y_t,t)^2\varphi''(Y_t)\right]. $$ Here the stochastic-integral term has expectation zero when its integrand is integrable; this is included in the stated regularity assumptions. Expressing the expectations as density integrals and integrating by parts gives $$ \frac{d}{dt}\int_{\mathbb R}\varphi(y)p_Y(y,t)\,dy = \int_{\mathbb R}\varphi(y) \left[-\frac{\partial}{\partial y}(\alpha p_Y) +\frac{1}{2}\frac{\partial^2}{\partial y^2}(\beta^2p_Y)\right]dy. $$ The boundary terms vanish by the assumed conditions. Since this identity holds for all such test functions, it is the forward equation in the distributional sense, and pointwise wherever the terms are sufficiently smooth. $\square$

Worked Example: Geometric Brownian Concentration

Suppose $dC_t=0.3C_t\,dt+0.4C_t\,dW_t$, with $C_0=1$ mg/L and $C_{\mathrm{ref}}=1$ mg/L. Find the log-coordinate SDE and the density at $t=2$ days.

Here $a(c,t)=0.3c$ and $b(c,t)=0.4c$. Therefore $$ \alpha=\frac{0.3c}{c}-\frac{(0.4c)^2}{2c^2} =0.3-0.08=0.22, \qquad \beta=\frac{0.4c}{c}=0.4. $$ Thus $dY_t=0.22\,dt+0.4\,dW_t$. Since $Y_0=0$, $Y_2\sim N(0.44,0.32)$: the mean is $0.22(2)=0.44$, and the variance is $0.4^2(2)=0.32$.

Consequently, $$ p_Y(y,2)=\frac{1}{\sqrt{2\pi(0.32)}} \exp\!\left[-\frac{(y-0.44)^2}{2(0.32)}\right], $$ and the concentration density is $$ p_C(c,2)=\frac{1}{c\sqrt{2\pi(0.32)}} \exp\!\left[-\frac{(\ln(c/C_{\mathrm{ref}})-0.44)^2}{2(0.32)}\right],\quad c>0. $$ The Itô correction $-0.08$ in the log drift is essential. The log drift is not $0.3$, even though the concentration drift is proportional to $C_t$.

Checks, Alternatives, and Limitations

There are two equivalent routes: transform the concentration density with the Jacobian, or transform the SDE with Itô’s formula and solve the resulting forward equation in $y$. They should agree when both are valid. The first route is especially direct when $p_C$ is already known; the second can simplify a model whose positive concentration has multiplicative noise. Neither route removes the need to check support, boundary behavior, and normalization.

Useful diagnostics include checking that $p_Y(y)=c\,p_C(c)$ at matching coordinates, verifying that the transformed density integrates to one, and checking that the support maps correctly: $c>0$ corresponds to $y\in\mathbb R$. A missing factor $1/c$ in the reverse conversion is a common error. Another is using the concentration drift divided by $c$ as the log drift while omitting the Itô correction. A further subtlety is that a density’s numerical value, and generally its mode, depends on the coordinate; the most likely value by density height need not map to the mode in the other coordinate.

The logarithmic transformation only represents strictly positive concentrations. If the law has an atom at zero, that atom cannot be represented by an ordinary density on finite $y$; it must be handled separately. The change-of-variables formulas also presume a one-to-one transformation over the support. For a different logarithm base, the Jacobian changes: for $Z=\log_{10}(C/C_{\mathrm{ref}})$, $dc/dz=(\ln 10)c$, so $p_Z(z)=(\ln 10)c\,p_C(c)$. Always use the derivative of the actual coordinate map.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $p_C(c)=1$ on $0<c<1$ and zero elsewhere, with $C_{\mathrm{ref}}=1$, what is $p_Y(y)$ and what is its support?
For a positive density $p_C$, write the formula for the probability $P(C>c_0)$ as an integral in log coordinates.
For $dC_t=\mu C_t\,dt+\sigma C_t\,dW_t$, what are the drift and diffusion coefficients of $Y_t=\ln(C_t/C_{\mathrm{ref}})$?
Why is a point mass at $C=0$ not represented by the ordinary log-concentration density?

Solutions. 1. Since $c=e^y$, $p_Y(y)=e^y$ when $y<0$, and zero otherwise. Its integral is $\int_{-\infty}^0 e^y\,dy=1$. 2. The event maps to $Y>\ln(c_0/C_{\mathrm{ref}})$, so its probability is $\int_{\ln(c_0/C_{\mathrm{ref}})}^\infty p_Y(y)\,dy$. 3. The drift is $\mu-\sigma^2/2$, and the diffusion coefficient is $\sigma$. 4. The logarithm of zero is not finite, so the atom lies outside the real-valued $Y$ coordinate and must be represented separately.

Log-Concentration Densities and the Jacobian

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