# Proof of Jensen's inequality for convexity

I am studying the Jensen inequality for convexity:

Let $$X$$ be a random variable. If $$g$$ is a convex function, then $$E(g(X)) \ge g(E(X))$$. If $$g$$ is a concave function, then $$E(g(X)) \le g(E(X))$$. In both cases, the only way that equality can hold is if there are constants $$a$$ and $$b$$ such that $$g(X) = a + bX$$ with probability $$1$$.

Then I am given a proof for this:

If $$g$$ is convex, then all lines that are tangent to $$g$$ lie below $$g$$ (see Figure 10.1). In particular, let $$\mu = E(X)$$, and consider the tangent line at the point $$(\mu, g(\mu))$$. (If $$g$$ is differentiable at $$\mu$$ then the tangent line is unique; otherwise, choose any tangent line at $$\mu$$.) Denoting this tangent line by $$a + bx$$, we have $$g(x) \ge a + bx$$ for all $$x$$ by convexity, so $$g(X) \ge a + bX$$. Taking the expectation of both sides,

$$E(g(X)) \ge E(a + bX) = a + bE(X) = a + b \mu = g(\mu) = g(E(X)),$$

as desired. If $$g$$ is concave, then $$h = -g$$ is convex, so we can apply what we just proved to $$h$$ to see that the inequality for $$g$$ is reversed from the convex case.

Lastly, assume that equality holds in the convex case. Let $$Y = g(X) - a - bX$$. Then $$Y$$ is a nonnegative r.v. with $$E(Y) = 0$$, so $$P(Y = 0) = 1$$ (even a tiny nonzero chance of $$Y > 0$$ occurring would make $$E(Y) > 0$$). So equality holds if and only if $$P(g(X) = a + bX) = 1$$. For the concave case, we can use the same argument with $$Y = a + bX - g(X)$$. $$\blacksquare$$ The last part of this proof is where I got confused:

Lastly, assume that equality holds in the convex case. Let $$Y = g(X) - a - bX$$. Then $$Y$$ is a nonnegative r.v. with $$E(Y) = 0$$, so $$P(Y = 0) = 1$$ (even a tiny nonzero chance of $$Y > 0$$ occurring would make $$E(Y) > 0$$). So equality holds if and only if $$P(g(X) = a + bX) = 1$$. For the concave case, we can use the same argument with $$Y = a + bX - g(X)$$.

It is my understanding that this last part of the proof is to show that the equality $$E(g(X)) = g(E(X))$$ only holds if there are constants $$a$$ and $$b$$ such that $$g(X) = a + bX$$ with probability $$1$$, which is why it begins with the assumption that equality holds in the convex case. However, there are a couple of points that I am confused about:

1. Why it is valid to assume that $$Y$$ is a nonnegative r.v. (although, it is clear to me why $$E(Y) = 0$$, based on the parts of the proof that came before this part).

2. Why is it that $$E(Y) = 0$$ implies that $$P(Y = 0) = 1$$?

3. Why is it that even a tiny nonzero chance of $$Y > 0$$ occurring would make $$E(Y) > 0$$?

Thank you.

1. Since by convexity $$g(X) \ge a + bX$$, so $$Y=g(X)-a-bX\geq 0$$.
2,3. The expectation property: if $$Y\geq 0$$ with probability $$1$$ and $$\mathbb E(Y)=0$$, then $$\mathbb P(Y=0)=1$$.
Proof. Consider for every $$x>0$$ the event $$\{Y > x\}$$ and find its probability. $$x\cdot\mathbb P(Y > x) \leq E (Y \mathbb 1_{Y>x}) \leq E(Y) = 0,$$ so $$\mathbb P(Y>x)=0$$ for all $$x>0$$. Next, $$Y>0$$ implies that there exists $$n=1,2,3,\ldots$$ such that $$Y>\frac1n$$. Then $$\mathbb P(Y>0) \leq \mathbb P\left(\bigcup_{i=1}^\infty \left\{Y>\frac1n\right\}\right)\leq \sum_{n=1}^\infty \mathbb P\left(Y>\frac1n\right) =0$$ Therefore $$1=\mathbb P(Y\geq 0)=\mathbb P(Y=0)+\underbrace{\mathbb P(Y>0)}_0 = \mathbb P(Y=0).$$
• But the author did not state that $Y\geq 0$ before claiming that $P(Y = 0) = 1$? Feb 23, 2020 at 9:27
• Look at the last yellow citation. > Let $Y = g(X) - a - bX$. Then $Y$ is a nonnegative r.v. with $E(Y) = 0$, so $P(Y = 0) = 1$