Random variable stochastic bigger than random variable

Question

I have a exercise, which I don't know how to show. It goes like, X is a continuous random variable with support $(-\infty,\infty)$. Consider the random variable $Y=X+\Delta$, where $\Delta$ is positive. I need to show that Y is stochastic bigger than X.

To show that Y is stochastic larger than X, I will use $P(Y>z)\geq P(X >z)$, which is the same as $F_Y(z) \leq F_X (z)$. I don't know how to include $Y=X+\Delta$, can anyone explain in details what to do, and which direction I should go?

Thanks in advance,

Answer 1

It suffices to show that $P(Y>z)-P(X>z) \geq 0$. Manipulating, $$P(Y>z) = P(X+\Delta>z) = P(X> z- \Delta)$$ So, if $f$ is your probability density function $$P(Y>z)-P(X>z) = \int_{z-\Delta}^{\infty} f(x) dx - \int_{z}^{\infty} f(x) dx$$ $$= \int_{z-\Delta}^{z} f(x) dx \geq 0$$ since $f$ is a pdf.

Answer 2

You will want to use the following theorem.

If $t_1\leq t_2$, then $F(t_1)\leq F(t_2)$

Proof: Note that $F(t_2)=P(X<t_2)=P\left(X<t_1\bigcup t_1\leq X<t_2\right)$ (to see this draw a picture of these intervals). Thus since $X<t_1$ and $t_1\leq X<t_2$ are disjoint we have finally that $F(t_2)=P(X<t_1)+P(t_1\leq X<t_2)\geq P(X<t_1)=F(t_1)$.

With this you can prove your result.

Answer 3

By definition probability measure is monotone. That is for $E_1\supset E_2\Longrightarrow$ $P(E_1)\geq P(E_2)$

Then from $P(Y>z)\geq P(X >z)$ we have $P(X+\Delta>z)\geq P(X >z)$ which is $P(X>z-\Delta)\geq P(X >z)$. We can see that $\{X>z-\Delta\} \supset \{X >z\}$, therefore the result follows.