91Ó°ÊÓ

Jensen's inequality states that for any convex function \(f\), and any random variable \(X\), the expected value satisfies \(E[f(X)] \geq f(E[X])\).

Let's apply Jensen's inequality to the exponential function \(f(x) = e^x\). Since the exponential function is convex, the inequality holds: \(E[e^X] \geq e^{E[X]}\).

By applying the AM-GM inequality, we have: \[ E\left[X^{2}\right]=E\left[X \cdot X\right] \geq E\left[\frac{X+X}{2}\right]^{2}=\left(\frac{E\left[2 X\right]}{2}\right)^{2}=\left(E[X]\right)^{2} \] Taking the square root of both sides, we get: \[ \left(E\left[X^{2}\right]\right)^{1 / 2} \geq E[X] \] We have proved the first part of the inequality.

To prove the remaining parts of the inequality, let's recall the definition of a convex function. A function \(g(x)\) is convex if for any two points \(x, y\) and \(0 \leq \lambda \leq 1\), \(g(\lambda x + (1 - \lambda) y) \leq \lambda g(x) + (1 - \lambda) g(y)\). Consider the function \(g(x) = x^{\frac{1}{n}}\), where n is a positive integer. We want to prove that \(g(x)\) is convex for \(x \geq 0\) in order to make use of Jensen's inequality. Let's show this by taking the second derivative of \(g(x)\). Calculate \(g'(x)\) and \(g''(x)\): \[ g'(x) = \frac{1}{n} x^{\frac{1}{n} - 1} \] \[ g''(x) = \frac{1}{n\left(n-1\right)} x^{\frac{1}{n} - 2} \] Since n is a positive integer, the second derivative of \(g(x)\) is non-negative for all \(x \geq 0\), and thus \(g(x)\) is convex for \(x \geq 0\). Now apply Jensen's inequality with \(g(x) = x^{\frac{1}{n}}\): \[ E\left[g(X)\right] = E\left[X^{\frac{1}{n}}\right] \geq g\left(E\left[X\right]\right) = \left(E\left[X\right]\right)^{\frac{1}{n}} \] Raising both sides of the inequality to the power of n gives: \[ \left(E\left[X^{\frac{1}{n}}\right]\right)^{n} \geq \left(E\left[X\right]\right) \] We have proved the second part of the inequality. Putting it all together: \[ E[X] \leq\left(E\left[X^{2}\right]\right)^{1 / 2} \leq\left(E\left[X^{3}\right]\right)^{1 / 3} \leq \cdots \]