existence of a RV with distribution given by a linear combination of other distributions

Question

Question:

Let $X$ and $Y$ be random variables defined on a $(\Omega,\mathfrak{F},\mathbb{P})$ probability space with distribution functions $F_X(t)$ and $F_Y(t)$, respectively.

(a) Show that for any $0< \alpha <1$, there exists a random variable $Z$ with the distribution function $$F_Z(t)=\alpha F_X(t)+(1-\alpha)F_Y(t).$$

(b) Assuming that $\mathbb{E}[|X|]< \infty$ and $\mathbb{E}[|Y|]< \infty$, compute $\mathbb{E}[Z]$.

Hint: Consider a new probability space $\Omega' =\Omega \times \{H,T\} $ and define a $\sigma$-algebra and probability measure appropriately.

Comments:

For part (a) I show that $F_Z(t)$ satisfies the three properties of the distribution function, namely being: non-strictly increasing, right-continuous and having limits $1$ and $0$ respectively at $+\infty$ and $-\infty$.

Then for part (b) by differentiating $$F_Z(t)=\alpha F_X(t)+(1-\alpha)F_Y(t)$$ we get $$f_Z(t)=\alpha f_X(t)+(1-\alpha)f_Y(t)$$, where $f_Z(t)$ is the density function and I use this fact and the linearity of integration to get that $\mathbb{E}[Z]=\alpha\mathbb{E}[X]+ (1-\alpha)\mathbb{E}[Y]$.

This solution that I wrote seems pretty easy and straightforward to me, but is it right? And if so why did the question give the hint, I can't see how can we use the hint, any hint on that? :D :D

Answer 1

You can define a probability measure on $\mathbb{P}'$ on space $\left(\Omega',\mathcal{F}'\right)$ by $\mathbb{P}'\left(A\times\left\{ H\right\} \right)=\alpha\mathbb{P}\left(A\right)$ and $\mathbb{P}'\left(A\times\left\{ T\right\} \right)=\left(1-\alpha\right)\mathbb{P}\left(A\right)$ for $A\in\mathcal{F}$. Then $Z:\Omega'\rightarrow\mathbb{R}$ defined by $\left(\omega,H\right)\mapsto X\left(\omega\right)$ and $\left(\omega,T\right)\mapsto Y\left(\omega\right)$ is $\mathcal{F}'$-measurable and has the mentioned cdf.

Showing that $F_Z$ has the characteristic properties of a cdf (as you did) is okay, but are you able to prove that for every cdf $F$ there exists a random variable $X$ having that cdf?

Since you have no further information about $F_X$ and $F_Y$ you should not use densities here: $$\int zd\left(\alpha F_{X}\left(z\right)+\left(1-\alpha\right)F_{Y}\left(z\right)\right)=\alpha\int zdF_{X}\left(z\right)+\left(1-\alpha\right)\int zdF_{Y}\left(z\right)=\alpha\mathbb{E}X+\left(1-\alpha\right)\mathbb{E}Y$$

Answer 2

You don't need to (actually shouldn't) assume they have a density (in fact, this is true even if they do not have a density) [ write the mean in terms of the CDF's of $X$ and $Y$ ].

You should be able to prove that prove for a r.v. $X$ with cdf $F_X(x)$,
$$E[X] = \int_0^\infty (1-F_X(x)) dx - \int_{-\infty}^0 F_X(x) dx$$ is true.

All you need to show is these 3 properties to show a function $F$ is a CDF:

• $F$ is nondecreasing
• $F$ is right continuous
• $\lim_{x\to \infty} F(x) = 1, \lim_{x \to -\infty} F(x) = 0$

Which are easy to check.

Existence of a random variable is also straight forward from a given CDF in general as given in this answer.