Constructing a PDF and CDF for a product of r.v.s

Question

If X~Expo(1) and S is a random sign (1 or -1 with each p=0.5) and S and are independent. How can I find the PDF of SX if I first should find the CDF of SX? I know that the resulting PDF should be somewhat similar to the Laplace PDF. I have been thinking about starting with something like:

$F(x)=\int_0^x{}S(t)e^{-t}dt$

But don't know if it's correct or how to treat S(t)... Any help would be appreciated!

Is the following correct?

For $x\geq0$, I get:

$F(x)=\frac{1}{2}(P(X\leq{}x)+1)=\frac{1}{2}(1+\int_0^xe^{-t}dt)=1-\frac{1}{2}e^{-x}$

And

For $x\leq0$, I get:

$F(x)=\frac{1}{2}(0+P(X\geq{}-x))=\frac{1}{2}(1-P(X\leq{}-x))=\frac{1}{2}(1-\int_0^{-x}e^{-t}dt)=\frac{1}{2}e^{x}$

Or have I misunderstood something?

A silly question I have is regarding the equation where $\frac{1}{2}$ is introduced. Is correct that we use that since S and X are independent then so are SX and S?

score 0 · Accepted Answer · answered Dec 03 '16 at 17:00

0

Hint: $$P(SX \le x) = P(SX \le x,S = 1) + P(SX \le x, S = -1) = \frac{1}{2}(P(X \le x) + P(X \ge -x))$$

Now consider the cases $x \ge 0$ and $x \le 0$ separately and remember that the PDF is the derivative of the CDF.

answered Dec 03 '16 at 17:00

Dominik

19,963

Thanks a lot for the quick reply! I will try go use your advice! – Nikolaj K Dec 03 '16 at 17:04
I have updated the question! If is possible I would be glad for help again:) – Nikolaj K Dec 03 '16 at 17:38
Your calculations are correct. Differentiating your result shows that the PDF ist given by $\frac{1}{2} e^{-|x|}$. – Dominik Dec 03 '16 at 17:42
1

Nevermind the new question. I have understood that it is just the law of total probabilities that has been applied in the first steps... – Nikolaj K Dec 03 '16 at 18:24

Constructing a PDF and CDF for a product of r.v.s

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