The formula presented in the image is:

$M_{Z}(\lambda) = \mathbb{E}[e^{\lambda Z}] \leq \exp\left(\frac{C^2 \lambda^2}{2}\right) \text{ for all } \lambda \in \mathbb{R},$

where $C \in \mathbb{R}$ depends on the distribution of the random variable $Z$. This inequality is commonly used in the context of deriving concentration inequalities, such as Chernoff bounds, by providing an upper bound on the moment generating function (MGF) of a random variable.

Step-by-Step Proof

Step 1: Definition of Moment Generating Function (MGF)

The moment generating function $M_Z(\lambda)$ of a random variable $Z$ is defined as:

$M_Z(\lambda) = \mathbb{E}[e^{\lambda Z}]$

Step 2: Apply Jensen's Inequality

Jensen's inequality states that for a convex function $f$ and a random variable $X$:

[ f(\mathbb{E}[X]) \leq \mathbb{E}[f(X)] ]

Since the exponential function $e^{\lambda Z}$ is convex, by Jensen's inequality:

[ M_Z(\lambda) = \mathbb{E}[e^{\lambda Z}] \leq \exp(\mathbb{E}[\lambda Z]) ]

However, for our bound, we need to consider the variance and higher-order moments, so we do not apply Jensen's inequality directly at this point.

Step 3: Taylor Expansion of Exponential Function

We consider the Taylor series expansion of the exponential function $e^{\lambda Z}$:

$e^{\lambda Z} = 1 + \lambda Z + \frac{\lambda^2 Z^2}{2} + \frac{\lambda^3 Z^3}{3!} + \dots$

Taking the expectation, we have:

[ \mathbb{E}[e^{\lambda Z}] = \mathbb{E}\left[1 + \lambda Z + \frac{\lambda^2 Z^2}{2} + \frac{\lambda^3 Z^3}{3!} + \dots \right] ]

Since $\mathbb{E}[Z] = 0$ (assuming zero mean for simplicity, which is common in many applications):

$M_Z(\lambda) = 1 + \frac{\lambda^2 \mathbb{E}[Z^2]}{2} + \text{higher-order terms}$

If we truncate the series to the quadratic term and assume that higher-order terms are negligible (for small $\lambda$), we obtain:

$M_Z(\lambda) \approx \exp\left(\frac{\lambda^2 \mathbb{E}[Z^2]}{2}\right)$

Step 4: Bounding the MGF

The above approximation suggests that:

$M_Z(\lambda) \leq \exp\left(\frac{C^2 \lambda^2}{2}\right)$

where $C^2$ can be interpreted as an upper bound on $\mathbb{E}[Z^2]$. The parameter $C$ depends on the distribution of $Z$ and represents a constant that bounds the second moment (variance) of $Z$.

Conclusion

Thus, the moment generating function $M_Z(\lambda)$ is bounded by $\exp\left(\frac{C^2 \lambda^2}{2}\right)$ for all $\lambda \in \mathbb{R}$. This bound is useful for deriving various concentration inequalities, such as Chernoff bounds, which are widely used in probability theory and statistics.

Further Details and Questions

Do you have any questions or would you like more details on specific steps of the proof?

Here are 8 related questions:

1. What assumptions are needed on the random variable $Z$ for the bound to hold?
2. How does the constant $C$ relate to the variance of $Z$?
3. Can this method be applied to non-zero mean random variables?
4. How does this bound lead to the Chernoff bound?
5. What happens if we do not truncate the Taylor series and include higher-order terms?
6. Can you provide an example of a random variable $Z$ where this bound is tight?
7. How is Jensen's inequality used differently in other types of bounds?
8. How would the bound change if $Z$ were not centered?

Tip: Understanding the derivation of moment generating functions and their bounds is crucial for applying various concentration inequalities in probability theory.