# PDF and normal distribution

1. May 1, 2014

### cwrn

1. The problem statement, all variables and given/known data
Two independent series of experiments are performed. The probability of a positive result (independent of eachother) in the respective series are given by p and q. Let X and Y be be the amount of experiments before the first negative result occur in the respective series.

1) Determine the probability functions for X and Y.

2) Let Z = min(X,Y), determine the distribution of Z by calculating the probability function. Give an interpretation of the answer. Why does Z have the distribution that was found?

3) Let Z ~ N(0,1) and let Y = Z2. Calculate the probability density function for Y.

4) Let X ~ N(μ, σ2) and let Y = X2. Calculate the probability density function for Y.

2. Relevant equations
Normal distribution:
\begin{align} f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \end{align}
3. The attempt at a solution

Since this is a problem from a previous exam in this course, I don't have an answer sheet, but on some parts I just need to know if I've done correctly or not.

1) Let X and Y be the random variables (discrete in this case). Since the experiments in a series are independent of eachother the probability function should look like this:
\begin{align} p_X(x) = p^x = Pr(X\ taking\ on\ value\ x) \\ p_Y(y) = q^y = Pr(Y\ taking\ on\ value\ y) \end{align}
(for x and y > 0)

The book I'm using is quite vague with the notation of probability functions, is the correct way to write it?

2) Does Z = min(X,Y) refer to an event where both experiments yields a negative result from the first attempt?
If that's the case the probability must be (1-p)(1-q)?

3) Z ~ N(0,1) means $\mu = 0, \ \sigma = 1$, which yields the probability density function
\begin{align} f_Z(z) = \varphi(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \end{align}
Since Y = Z2, it must mean that y = z2. So the solution is given simply by substituting z?

4) I assume this is almost like 3), but with different values?

Last edited: May 1, 2014
2. May 1, 2014

### Ray Vickson

The X, Y and Z in parts 1--2 have no relation at all to the X,Y and Z in parts 3--4.

In part 2, Z is the smaller of the two random variables X and Y; there is no assumption about the values of X and Y. For example, you might observe X = 47 and Y = 63, and in that case you observe Z = 47. Basically, you are asked to determine the distribution of the smaller of two independent random variables.

In part 3, you cannot just put y = z^2; you must use the standard "change of variable" formulas for probability densities. If you have not seen these before, start by finding the cdf $F_Y(y)$ of Y on its domain $y \geq 0$: $$F_Y(y) = P\{ Y \leq y \} = P\{ Z^2 \leq y \} = P\{|Z| \leq \sqrt{y} \} = P\{ -\sqrt{y} \leq Z \leq \sqrt{y} \}$$
Now get the density as $f_Y(y) = dF_Y(y)/dy$. Do something similar for part 4.

Last edited: May 1, 2014
3. May 1, 2014

### cwrn

Alright, according to the definition, the distribution should be
\begin{align} F_Z(u) = 1-(1-F_X(u))(1-F_Y(u)) = 1-P(X>u)P(Y>u) \end{align}
where $F_Y(u)=\sum_{k:k\leq u}q^k$ and $F_X(u)=\sum_{k:k\leq u}p^k$.

Last edited: May 1, 2014
4. May 1, 2014

### cwrn

I gave part 3 another try using the so called "change of variable" technique. Although, instead of finding and taking the derivative of the CDF I used the definition of normal distribution (pdf) and just substituted z. I set $z=g(y)=\sqrt{y} \Rightarrow \frac{dg}{dy}=\frac{1}{2\sqrt{y}}$. Substituting z gives
\begin{align} f_Y(y) = f_Z(g(y)) \left|{\frac{dg}{dy}}\right| = \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\left|{\frac{1}{2\sqrt{y}}}\right|. \end{align}
Is this correct?

Last edited: May 1, 2014
5. May 1, 2014

### Ray Vickson

Not quite. The method you use can give trouble when you are dealing with a non-monotone function of a random variable, and in this case $X^2$ is non-monotone. That is precisely why I suggested the alternative method, which is always 100% correct, as it forces you to be careful.
$$P(Y \leq y) = \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}\, dx = 2\int_0^{\sqrt{y}}\frac{1}{\sqrt{2 \pi}} e^{-x^2/2}\, dx$$
Thus
$$f_Y(y) = \frac{d}{dy} 2\int_0^{\sqrt{y}}\frac{1}{\sqrt{2 \pi}} e^{-x^2/2}\, dx = \frac{2}{\sqrt{2 \pi}} e^{-y/2} \frac{d \sqrt{y}}{dy} = \frac{1}{\sqrt{2 \pi}} \frac{e^{-y/2}}{\sqrt{y}}$$
This integrates to 1, so yours will integrate to 1/2.

6. May 2, 2014

### cwrn

Of course. It makes a lot more sense to look at the substitution in an integration. Thanks for the help. Will try part 4) and post the result later.

Edit: What do you mean by "yours will integrate to 1/2"? Do I need to change the limits according to z(y)?

Last edited: May 2, 2014
7. May 2, 2014

### Ray Vickson

I mean that your "f(y)" has the wrong normalization. Your answer is (1/2)f(y), not f(y) itself. Compare yours and mine very carefully!

8. May 2, 2014

### cwrn

I see, I must've misinterpreted, my bad! Thanks again.

9. May 2, 2014

### cwrn

Using the same method in 4) as in 3) gives me
\begin{align} f_Y(y) = \frac{1}{\sqrt{2\pi y\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} \end{align}

Last edited: May 2, 2014
10. May 2, 2014

### Ray Vickson

This is just about as wrong at it could possibly be. Go back to square one, and proceed carefully.