Prove Jensen's Inequality: Var[X] ≥ (E[X])^2

[jetoso]

The variance can be written as Var[X]=E[X^2]-(E[X])^2. Use this form to prove that the Var[X] is always non-negative, i.e., show that E[X^2]>=(E[X])^2.
Use Jensen's Inequality.

Any sugestions? I just tried to prove that a function g(t) is continuous and twice differentiable, such that g''(t) > 0 which must imply it is convex.
Then, I am stuck with the proof.

 

[mathman]

I am not sure how to use Jensen's inequality. However by using the relationship:

E((X-E(X))²)=E(X²)-(E(X))², the result is obvious.

 

[tacman]

Sure, I can help with the proof. First, let's define the function g(t) as g(t) = E[(X-t)^2]. This function represents the expected value of the squared deviations of X from t. We can rewrite this as g(t) = E[X^2] - 2tE[X] + t^2.

Now, let's take the second derivative of g(t) with respect to t:

g''(t) = 2E[X] - 2E[X] = 2(E[X] - E[X]) = 2Var[X]

Since Var[X] ≥ 0, we can conclude that g''(t) ≥ 0, which means that g(t) is a convex function. This is because the second derivative of a convex function is always non-negative.

Now, let's use Jensen's Inequality, which states that for any convex function g(t) and any random variable X, we have E[g(X)] ≥ g(E[X]). Applying this to our function g(t), we have:

E[(X-E[X])^2] ≥ (E[X]-E[X])^2

Simplifying, we get:

E[(X-E[X])^2] ≥ 0

But this is just the definition of Var[X]. Therefore, we have shown that Var[X] ≥ 0, or in other words, Var[X] is always non-negative.

To summarize, we used Jensen's Inequality and the convexity of the function g(t) to prove that Var[X] is always non-negative. This implies that Var[X] ≥ (E[X])^2, which is the desired result.