# Are these two inequalities are equivalent?

Let's assume that $I_j \in \mathcal{J}$, where $\mathcal{J}$ is a set of images that are correctly classified and $p(I)$ is the output probability distribution of the used underlying model. Out of $\mathcal{J}$, we select $\hat{I}_{\!j^*}$ according to a well defined metric according to the inequality shown in (2). Given that $p(\hat{I}_{\!j^*} | \hat{y}= y) \leq p(\hat{I}_{\!j} | \hat{y}= y), $ where $y$ is the Ground Truth label. However, for $p(\hat{I}_{\!j^*} | \hat{y}= y) > p(\hat{I}_{\!j^*} | \hat{y}\neq y)$ and $p(I_j| \hat{y}= y) < p(I_j | \hat{y}\neq y)$.

How could we show that the first inequality shown in (1) is equivalent to the second inequality shown in (2)?

$$ \sum p(\hat{I}_{\!j^*})\log p(\hat{I}_{\!j^*}) \leq \sum p(\hat{I}_{\!j}) \log p(\hat{I}_{\!j}) \tag{1} $$

$$ \sum \log p(\hat{I}_{\!j^*}) \leq \sum \log p(\hat{I}_{\!j}) \tag{2} $$**A possible arugment that I believe might solve the problem is the following:**

We can observe that the difference between both inequalities lies in the fact that (1) can be viewed as a scaled version of (2), where the probabilities are multiplied by their logarithms. Considering the monotonically increasing nature of the logarithmic function, we can roughly establish the equivalence of these two inequalities.

I have tested it numerically, and both inequalities are equivalent, but I cannot prove it mathematically.

- unanswered
- 148 views
- Pro Bono

### Related Questions

- Figuring out the maths for the probability of two adopted teens randomly being matched as pen pals in 2003
- Probability of picking a red ball
- Probabilities
- Two exercises in complex analysis
- real analysis
- Limit of an Integral of a $C^\infty$-Smooth Function with Compact Support
- $\textbf{I would like a proof in detail of the following question.}$
- Check if problems are correct

Questions at this level should come with a bounty.