Suppose $X_1$ and $X_2$ are $N(0,1)$ random variables such that $X_2=-X_1$ if $|X_1|<1$ and $X_2=X_1$ otherwise. Obtain the cumulative distribution function of $X_1+X_2$ represented in the form(s) of cumulative distribution function of a standard normal variable.

I get $P(X_1+X_2<z)=\Phi (z/2)$ if $|z|>2$. But what about the case $|z|<2$? I would like to get convinced that my working is right. What I think will happen is, if $-2<z<2$ then $P(X_1+X_2<z)=P(X_1+X_2<-2)+P(-2<X_1+X_2<z)=\Phi (-1)+P(-2<0<z)=\Phi (-1)+0.5$

This is because $0$ is either less than or greater than $z$ which is a real number. So $P(-2<0<z)=P(0<z)=0.5$.

Is the reasoning correct? Thanks in advance.

  • $\begingroup$ Consider: What are $F(-2)$ and $F(2)$? What's $P(X_1+X_2=0)$? What's $P(-2<X_1+X_2<0)$? What's $P(0<X_1+X_2<2)$? If you can answer those, you should be able to write down the whole CDF by inspection. $\endgroup$ – Glen_b Nov 2 '14 at 9:47
  • $\begingroup$ What do you denote by $F$, please state more clearly. $P(X_1+X_2=0)=P(-1<X_1<1)=2\Phi (1)-1$. $P(-2<X_1+X_2<0)=$ my doubt. Same goes for $P(0<X_1+X_2<2)$. But as I said, the event $X_1+X_2$ between 0 and 2 occurs ONLY when $X_1+X_2=0$ i.e. the probability of this event is $P(-1<X_1<1)$ or less. Need some help here. $\endgroup$ – Landon Carter Nov 2 '14 at 9:55
  • $\begingroup$ $F$ is the cdf of $X_1+X_2$. You should have no doubt what $P(X_1+X_2=0)$ is. Under what circumstances is $X_2=-X_1$ ... (which is the same as $X_1+X_2=0$)? $\endgroup$ – Glen_b Nov 2 '14 at 10:11
  • $\begingroup$ $P(X_1+X_2=0)=2\Phi (1)-1$ : is this not correct? $\endgroup$ – Landon Carter Nov 2 '14 at 10:17
  • $\begingroup$ So can you figure out the other quantities I mentioned now? $\endgroup$ – Glen_b Nov 2 '14 at 11:05

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