You're right to point out that $\gamma$ and $(1-\gamma)$ appear in the denominator in the expression $\frac{P}{\gamma R_1 + (1-\gamma) R_2}$, which complicates directly applying the convexity property as initially described. Let's revise the approach to address this correctly.

Revised Proof

To prove the inequality:

$\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx \geq P$

Given the two inequalities: $\int \min \left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) dx \geq P$ $\int \min \left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right) dx \geq P$

Step 1: Define an auxiliary function

Define: $f_\gamma(x) = \gamma f_1(x) + (1-\gamma) f_2(x)$ where $f_1(x) = \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right)$ and $f_2(x) = \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right)$.

Step 2: Use the convexity of the auxiliary function

We will examine the function inside the integral for $\gamma$ between 0 and 1: $\min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right)$

Although directly applying convexity to the minimum function isn't straightforward because of the $\gamma$ and $(1-\gamma)$ in the denominator, we can approach it by considering that the entire expression is a convex combination of two terms, and then use the properties of the min function:

$\min(a, b) \geq \min(c, d) \quad \text{if} \quad a \geq c \ \text{and} \ b \geq d$

Step 3: Convexity of the $\frac{1}{\gamma R_1 + (1-\gamma) R_2}$

Now, consider: $\frac{1}{\gamma R_1 + (1-\gamma) R_2} \leq \frac{\gamma}{R_1} + \frac{1-\gamma}{R_2}$

Using this inequality, we compare: $\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x) \geq \gamma \frac{P}{R_1} p_X(x) + (1-\gamma) \frac{P}{R_2} p_X(x)$

Hence, we can assert: $\min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) \geq \min \left(\gamma \frac{P}{R_1} p_X(x) + (1-\gamma) \frac{P}{R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right)$

Step 4: Integrate the revised inequality

Finally, integrate both sides: $\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx$

Using the convexity argument with integration: $\geq \gamma \int \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) dx + (1-\gamma) \int \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right) dx$

Substitute the given inequalities: $\geq \gamma P + (1-\gamma) P = P$

Thus, we conclude: $\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx \geq P$

This completes the proof.

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Would you like to discuss any specific steps in this revised proof or any other aspects of the problem?

Related Questions:

1. How does Jensen's inequality apply to convex functions in probability theory?
2. Can you further explain the convexity of the function $\frac{1}{\gamma R_1 + (1-\gamma) R_2}$?
3. What are the implications of this proof in information theory?
4. How might the proof change with different forms of probability density functions $p_X$?
5. What is the significance of using the min function in this context?

Tip:

When dealing with expressions that involve denominators in convex combinations, carefully consider the convexity or concavity of the involved functions, as they can often simplify the proof or analysis.