I am using Jensen's inequality and conditional expectation to prove the following inequality:

Let $\lambda_i$ be real for $i\in \{1,2,...,M\}$ and $\bar{\lambda}=\frac{\sum_{i=1}^M\lambda_i}{M}$. Let $X_i$, $i\in \{1,2,...,M\}$ be a set of real i.i.d random variables, then we have \begin{align} \mathbb{E}\left(f\left(\sum_{i=1}^M \bar{\lambda} X_i\right)\right) \le \mathbb{E}\left(f\left(\sum_{i=1}^M \lambda_i X_i\right)\right), \end{align} where $f(\cdot)$ is a convex function. The equal sign holds when $\lambda_{i}=\bar{\lambda}$ for all $i\in \{1,2,...,M\}$.

And my proof is given as:

Let \begin{align} X=\sum_{i=1}^M\bar{\lambda}X_i, \end{align} \begin{align} W=\sum_{i=1}^M\lambda_iX_i, \end{align} and \begin{align} Z=X-W. \end{align} By the symmetric property of $X$, the conditional random variables $X_i|X$, $i\in \{1,...,M\}$, are identically distributed, which implies \begin{align} \mathbb{E}(X_1|X)=\mathbb{E}(X_2|X)= \cdots = \mathbb{E}(X_M|X). \end{align} Therefore \begin{align} \mathbb{E}(Z|X)&=\sum_{i=1}^M \bar{\lambda}\mathbb{E}(X_i|X)-\sum_{i=1}^M \lambda_{i}\mathbb{E}(X_i|X)\\ &=\mathbb{E}(X_1|X)\left(\sum_{i=1}^M \bar{\lambda}-\sum_{i=1}^M \lambda_{i}\right)=0. \end{align} Since $f(\cdot)$ is convex, by Jensen's inequality we have \begin{align} \mathbb{E}(f(X-Z)|X) &\ge f(\mathbb{E}((X-Z)|X)) \\ &=f(X-0)=f(X) \end{align} Therefore \begin{align} \mathbb{E}(f(W))=\mathbb{E}(\mathbb{E}(f(X-Z)|X))\ge \mathbb{E}(f(X)). \end{align}

I think the proof is right, but not 100% sure, can someone give me a judgement of this proof?


  • $\begingroup$ your question is about a specific problem using Jensen's inequality, right? Then your title is way to general considering the focused scope of your question. Please change your title. $\endgroup$ Nov 17 '14 at 15:33

It looks all right to me.

(I would love to found some small error, or to have some improvement to suggest, or -damn- anything interesting to say so that the MSE system allows me to post this as answer without violating the "more than 30 characters" rule [well, that's done] and so that the downvoters deem this as a valid answer - but so is life. The question body is completely correct, and this answer is also - especially when one disregards this paragraph)

  • $\begingroup$ This was posted as a comment to the OP 23 hours ago. $\endgroup$ Aug 15 '13 at 18:41
  • $\begingroup$ Yes. I just reread it and I got a bit more confident. So? Why the downvote? What do you suggest? To leave the question without answer? $\endgroup$
    – leonbloy
    Aug 15 '13 at 18:43
  • $\begingroup$ The question was already answered, in fact by you. You don't need to repost. That's all I am saying. Please don't take offense. $\endgroup$ Aug 15 '13 at 18:47
  • $\begingroup$ I take offense with the unjustied downvotes, because I guess they presume vote-whoring on my part. I'm simply moving the comment as answer so that this question does not look as "pending" anymore. That's the only answer that a "verify this proof" for me" can ellicit (if the proof is correct, of course). $\endgroup$
    – leonbloy
    Aug 15 '13 at 18:52
  • 2
    $\begingroup$ @leonbloy I think moving a comment, which answers the question fully, to an answer (so that it moves this question off of the unanswered queue) is a good thing. +1 thankyou. $\endgroup$
    – Dan Rust
    Aug 15 '13 at 18:57

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