The arrival of photons at a pixel in an image sensor is a Poisson distributed random variable such that the input can be modeled as a Poisson r.v. $X\sim \mathrm{Poisson}(\lambda)$.

Since the input is Poisson, the mean and variance are equal such that

\begin{equation} \frac{\mathbb{E}[X]}{\mathrm{Var}[X]}=1 \end{equation}

Now when the photon input is passed through a linear image sensor (camera) to produce a digital output, we can treat this as a linear transformation of $X$ such that the output, $Y$ is $Y=X/g$.

In the case of this linear sensor, I can extract the `conversion gain', i.e. number of photons required to produce a digital output of one, represented as $g$ in units of (photons/digital#), as

\begin{equation} \frac{\mathbb{E}[Y]}{\mathrm{Var}[Y]}=\frac{\mathbb{E}[X/g]}{\mathrm{Var}[X/g]}=\frac{\frac{1}{g}\mathbb{E}[X]}{\frac{1}{g^2}\mathrm{Var}[X]}=g \end{equation}

However, now consider a sensor where the conversion gain depends linearly on the input, e.g. $Y=X/(aX+b)$ where $a>0$ and $b>0$. This means that the gain is an increasing function of signal $g(x)=ax+b$.

In the case of this non-linear sensor, the gain can no longer be found from the ratio of the mean to variance at the output

\begin{equation} \frac{\mathbb{E}[Y]}{\mathrm{Var}[Y]}\neq g(x) \end{equation}

In fact, the measured conversion gain is found to be larger than the actual conversion gain for any input signal level.

\begin{equation} \frac{\mathbb{E}[Y]}{\mathrm{Var}[Y]}> g(x) \end{equation}

Part of the explanation for this is Jensen's inequality which states that for a increasing concave transformation of some random input $X$, i.e. $Y=f(X)$:

\begin{equation} \mathbb{E}[Y]=\mathbb{E}[f(X)]\leq f(\mathbb{E}[X]) \end{equation}

In my case $Y=X/(aX+b)$ is in fact a increasing concave function meaning that the measured mean at the output is less than the transformed mean of the input. Since we know the measured gain at the output is overestimated and the measured mean is underestimated, this implies that the measured variance is even more underestimated than the mean.

How can I prove this or mathematically write this? Is there a generalization of Jensen's inequality for the variance? Can I show exactly why the gain is overestimated in this example?

  • $\begingroup$ If $g$ is a function of $X$, it has just changed dramatically from a constant to a random variable, i.e. a non-constant function. Then, how could it remain equal to the ratio of two constant quantities (mean/variance)? I don't see room for a "deeper" understanding. $\endgroup$ Commented Jan 20, 2017 at 21:21

1 Answer 1


There is no relation between the two quantities $f(\text{Var}[X])$ and $\text{Var}[f(X)]$ for concave $f$. Here are the examples to demonstrate this:

  • Ex 1: Suppose random variable $X$ has the pmf: $p_X(0) = \frac{1}{2}$ and $p_X(4) = \frac{1}{2}$, and $f(x) = \sqrt{x}$. We get $\text{Var}[f(X)] = 1$ and $f(\text{Var}[X]) = f(4) = 2$. So, $\text{Var}[f(X)] < f(\text{Var}[X])$.
  • Ex 2: Suppose random variable $X$ is same as before i.e. it has the pmf: $p_X(0) = \frac{1}{2}$ and $p_X(4) = \frac{1}{2}$, but $f$ changed to $f(x) = \sqrt{x} - 100$. Note that $\text{Var}[f(X)] = 1$ still, but now $f(\text{Var}[X]) = f(4) = 2-100=-98$. So, $\text{Var}[f(X)] > f(\text{Var}[X])$.
  • 1
    $\begingroup$ That's a good point. In my example I was not as specific as I should have been. I am interested in the case of increasing, positive, concave functions. Does a relationship exist under that class of functions? $\endgroup$ Commented Jan 23, 2017 at 15:57
  • $\begingroup$ Ex 3: Suppose $X$ has a pmf: $p_X(0) =p_X(4) = \frac{1}{2}$, but $f(x)$ is $f(x) = 100\sqrt{x}$. $\text{Var}[f(X)]$ is equal to $10000$, but $f(\text{Var}[X]) = 200$. Therefore, $\text{Var}[f(X)] > f(\text{Var}[X])$. Check that in examples 1 and 3, $f$ is increasing positive and concave. $\endgroup$
    – Amit
    Commented Jan 23, 2017 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.