How can I compute $\int_{-\infty}^\infty I(t) \lambda (dt)$ and $\int_{-\infty}^\infty I^2(t) \lambda (dt)$?

Question

The Question is,

Compute $E(X)$, $E(X^2)$ and $Var(X)$, where the law of X is given by

$L(X)= {1 \over 2} \delta_1 + {1\over 2} \lambda$, where $\lambda$ is Lebesgue measure on $[0,1]$.

The Solution is, when $I(t)=t$,

$E_P(X)= {1 \over 2} \int_{-\infty}^\infty I(t) \delta_1 (dt) + {1 \over 2} \int_{-\infty}^\infty I(t) \lambda (dt)= {1 \over 2}(1) + {1\over2}({1\over2})={3\over4}$

$E_P(X^2)={1 \over 2} \int_{-\infty}^\infty I^2(t) \delta_1 (dt) + {1 \over 2} \int_{-\infty}^\infty I^2(t) \lambda (dt)= {1 \over 2}(1) + {1\over2}({1\over 3})={2\over3}$

$Var(X)=E_P(X^2)-E_P^2(X)={5 \over 48}$

But the solution is above my head.

How can I compute $\int_{-\infty}^\infty I(t) \lambda (dt)$ and $\int_{-\infty}^\infty I^2(t) \lambda (dt)$?

Answer 1

This is a consequence of the fundamental theorem of calculus, see https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements

$$\int_{-\infty}^\infty I(t)\lambda(dt) = \int_0^1t\, dt=\left[\frac{1}{2}t^2\right]^1_0=\frac{1}{2}.$$

Just remember to insert the definitions of $I(t)$ and $\lambda$ and you obtain a Riemannian integral.

Answer 2

Here what you call "the Lebesguemeasure on $[0,1]$" is actually the measure prescribed by $A\mapsto\lambda (A\cap[0,1])$ where $\lambda$ denotes the Lebesguemeasure.

Then: $$\mathsf Ef(X)=\frac12\int f(t)\delta_1(dt)+\frac12\int f(t)\lambda(dt)=\frac12f(1)+\frac12\int_{[0,1]} f(t)dt$$

Can you find $\int_{[0,1]} tdt$ and $\int_{[0,1]} t^2dt$ yourself?