# Proof by Induction (Urgent hint required)

• Nov 24th 2008, 05:38 AM
slevvio
Proof by Induction (Urgent hint required)
Let X be a discrete random variable and $g: R_X \rightarrow \mathbb{R}$ a continuous function, which is convex, i.e. for all $x_1,x_2 \in R_X$ and $\lambda \in (0,1)$

$g(\lambda x_1 + (1-\lambda)x_2 \leq \lambda g(x_1) + (1-\lambda) g(x_2).$

(a) For $\lambda_i \geq 0$ with $\sum_{i=1}^{n} \lambda_i = 1$ show that $g(\sum_{i=1}^{n} \lambda_i x_i ) \leq \sum_{i=1}^{n} \lambda_i g(x_i)$

I have let P(n) denote the above statement and shown that it holds for n = 1, but I am finding it hard to progress much further. I really need help with this and any hints or tips would be appreciated.
• Nov 25th 2008, 04:41 AM
CaptainBlack
Quote:

Originally Posted by slevvio
Let X be a discrete random variable and $g: R_X \rightarrow \mathbb{R}$ a continuous function, which is convex, i.e. for all $x_1,x_2 \in R_X$ and $\lambda \in (0,1)$

$g(\lambda x_1 + (1-\lambda)x_2 \leq \lambda g(x_1) + (1-\lambda) g(x_2).$

(a) For $\lambda_i \geq 0$ with $\sum_{i=1}^{n} \lambda_i = 1$ show that $g(\sum_{i=1}^{n} \lambda_i x_i ) \leq \sum_{i=1}^{n} \lambda_i g(x_i)$

I have let P(n) denote the above statement and shown that it holds for n = 1, but I am finding it hard to progress much further. I really need help with this and any hints or tips would be appreciated.