Probability Functions and Distributions for Independent Series of Experiments

[cwrn]

Homework Statement

Two independent series of experiments are performed. The probability of a positive result (independent of each other) 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.

Homework Equations

Normal distribution:
$$
\begin{align}
f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
\end{align}
$$

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 each other 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?

 

[Ray Vickson]

Homework Statement

Two independent series of experiments are performed. The probability of a positive result (independent of each other) 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.

Homework Equations

Normal distribution:
$$
\begin{align}
f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
\end{align}
$$

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 each other 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?

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.

 

[cwrn]

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.

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.

 

[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?

 

[Ray Vickson]

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?

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<br /> = 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<br /> = \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.

 

[cwrn]

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<br /> = 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<br /> = \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.

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)?

 

[Ray Vickson]

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)?

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!

 

[cwrn]

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!

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

 

[cwrn]

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

 

[Ray Vickson]

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

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