Question

Let \(\varphi(\xi)\) be a symmetric function, nondecreasing in \(|\xi|\), with \(\varphi(0)=0\), and such that \(\left\\{\varphi\left(\dot{Y}_{j}\right)\right\\}_{j=0}^{n}\) is a submartingale. Fix \(0=u_{0} \leq u_{1} \leq \cdots \leq u_{n}\). Show that $$ \operatorname{Pr}\left\\{\left|Y_{j}\right| \leq u_{j} ; 1 \leq j \leq n\right\\} \geq 1-\sum_{j=1}^{n} \frac{E\left[\varphi\left(Y_{j}\right)\right]-E\left[\varphi\left(Y_{j-1}\right)\right]}{\varphi\left(u_{j}\right)} $$ (If \(\varphi(\xi)=\zeta^{2}, u_{1}=\cdots=u_{n}=\lambda\), we obtain Kolmogorov's inequality.)

Short answer

Therefore, the proof of the given inequality is established as \( \operatorname{Pr}\left\{\left|Y_{j}\right| \leq u_{j} ; 1 \leq j \leq n\right\} \geq 1-\sum_{j=1}^{n} \frac{E\left[\varphi\left(Y_{j}\right)\right]-E\left[\varphi\left(Y_{j-1}\right)\right]}{\varphi\left(u_{j}\right)} \)

Step-by-step solution

Write out the properties of the function and submartingales

From the given conditions, it is known that \( \varphi(\xi) \) is symmetric and nondecreasing in \( |\xi| \), with \( \varphi(0) = 0 \). Also, given that \( \left\{\varphi\left(\dot{Y}_{j}\right)\right\}_{j=0}^{n} \) is a submartingale, we can say that the expected value of \( \dot{Y}_{j} \) conditioned on the previous \(\dot{Y}_{j-1}\) increases or stays constant, that is, \( E\left[\varphi\left(\dot{Y}_{j}\right) | \dot{Y}_{j-1}\right] \geq \varphi\left(\dot{Y}_{j-1}\right) \).

Analyze the Probability

We need to consider the probability \( Pr\left\{\left|Y_{j}\right| \leq u_{j} ; 1 \leq j \leq n\right\} \). We can use the indicator function to represent this event as \( 1_{\left\{ \left|Y_{j}\right| \leq u_{j}\right\}} \). Notice that the function \(\varphi(\xi)\) is non-decreasing, so we can say \( \varphi(\left|Y_{j}\right|)\) is less than or equal to \( \varphi(u_{j}) \) when \( \left|Y_{j}\right| \leq u_{j} \) i.e. \( \varphi(\left|Y_{j}\right|) 1_{\left\{ \left|Y_{j}\right| \leq u_{j}\right\}} \leq \varphi(u_{j})1_{\left\{ \left|Y_{j}\right| \leq u_{j}\right\}} \)

Apply the submartingale property

Take the expectation on both sides and apply the submartingale property: \( E\left[\varphi(\left|Y_{j}\right|)\right]-E\left[\varphi(\left|Y_{j-1}\right|)\right] \leq E\left[\varphi\left(u_{j}\right) 1_{\left\{ \left|Y_{j}\right| \leq u_{j}\right\}} \right] - E\left[\varphi\left(u_{j-1}\right) 1_{\left\{ \left|Y_{j-1}\right| \leq u_{j-1}\right\}} \right] \). Now, if we denote \( p_{j} = Pr\left\{ \left|Y_{j}\right| \leq u_{j} \right\} \), we can relate the inequality to the probability.

Relate inequality to probability

If we multiply both sides by \( \frac{1}{\varphi\left(u_{j}\right)} \) we get, \( \frac{E\left[\varphi\left(Y_{j}\right)\right]-E\left[\varphi\left(Y_{j-1}\right)\right]}{\varphi\left(u_{j}\right)} \leq p_{j} - p_{j-1} \). Then sum over \( j=1,2,...,n \), and we obtain \( 1- \sum_j \frac{E\left[\varphi\left(Y_{j}\right)\right]-E\left[\varphi\left(Y_{j-1}\right)\right]}{\varphi\left(u_{j}\right)} \leq p_{n} \), which leads to the desired inequality.