Random variable and probability density function

[PainterGuy]

Homework Statement

The problem is about continuous random variable and probability density function.

Relevant Equations

Please check the posting.

Hi,

I was trying to solve the attached problem which shows its solution as well. I cannot understand how and where they are getting the equations 3.69 and 3.69A from.

Are they substituting the values of θ₁ and θ₂ into Expression 1 after performing the differentiation to get equations 3.70 and 3.71?

As you can see that the solution is there but I need to understand it. I'd really appreciate if you could help me with it. Thank you!

Hi-resolution image copy: https://imagizer.imageshack.com/img923/1491/lXm2aZ.jpg

 

[anuttarasammyak]

I am not familiar with the notation
$$f_y(y)$$
I see it is a function with variable y, but what does y in $f_y$ mean ? It seems $f_y(y)$ is defined somewhere before the problem in your text.

 

[PainterGuy]

I am not familiar with the notation
$$f_y(y)$$
I see it is a function with variable y, but what does y in $f_y$ mean ? It seems $f_y(y)$ is defined somewhere before the problem in your text.

It means probability density function of 'f()' where "Y" is a random variable and "y" is a dummy variable, i.e. f_Y(y).

 

[anuttarasammyak]

Thanks. So I observe correspondence of areas for distribution of y and $\theta$ as attached.

 

[PainterGuy]

My apologies but I still don't get how and where they are getting the equations 3.69 and 3.69A from.

 

[anuttarasammyak]

$$f_y(y)dy = f_\theta(\theta) d\theta|_{\theta_1}+f_\theta(\theta) d\theta|_{\theta_2}$$ or more strictly on sign
$$f_y(y)|dy| = f_\theta(\theta) |d\theta||_{\theta_1}+f_\theta(\theta) |d\theta||_{\theta_2}$$
Divide the both sides by |dy|.

 

[PainterGuy]

What is $f_\theta(\theta) d\theta$ and where does this come from? I don't see its connection with the original function y=cos(α+θ). Could you please help?

 

[anuttarasammyak]

$$f_\theta(\theta)\ d\theta|_{\theta_1}$$
is probability of $\theta$ to take value $[\ \theta_1-\frac{1}{2} d\theta,\ \theta_1+ \frac{1}{2}d\theta\ ]$
It comes from the graph of y-$\theta$ correspondence as I have attached.

 

[PainterGuy]

$$f_\theta(\theta)\ d\theta|_{\theta_1}$$
is probability of $\theta$ to take value $[\ \theta_1-\frac{1}{2} d\theta,\ \theta_1+ \frac{1}{2}d\theta\ ]$
It comes from the graph of y-$\theta$ correspondence as I have attached.

Thank you but, honestly speaking, I'm still having a hard time. I think I should give it few hours, perhaps it'd make sense.

 

[vela]

If you have a random variable $X$, you can have a function of that variable $g(X)$ which defines a new random variable $Y=g(X)$. $X$ has a pdf $f_X(x)$; similarly, the variable $Y$ has a pdf $f_Y(y)$.

Assume for a moment that $g(x)$ monotonically increases. The function $g$ maps the interval $(x,x+dx)$ to the interval $(y,y+dy)$, so the probability that $X$ is between $x$ and $x+dx$ should be equal to the probability $Y$ is between $y$ and $y+dy$. In terms of the pdfs, you have
$$f_X(x)\,dx = f_Y(y)\,dy.$$ Since $dy = g'(x)\,dx$, you get $f_Y(y) = f_X(x)/g'(x)$.

If $g$ were a decreasing function, however, we'd run into a problem as the pdfs need to be non-negative. In that case, we should have $f_Y(y) = -f_X(x)/g'(x)$ instead. To cover both cases, we use the absolute value of $g'(x)$.

Finally, if we drop the requirement that $g$ is a one-to-one function, then it's possible that more than one value of $x$ maps to a particular value of $y$. In that case, we need to sum the probabilities to get
$$f_Y(y)\,dy = \sum_{g(x_i)=y} f_X(x_i)\,dx \quad \Rightarrow \quad f_Y(y) = \sum_{g(x_i)=y} \frac{f_X(x_i)}{\lvert g'(x_i) \rvert}.$$ This formula applied to this particular problem is where equation 3.69 comes from.

 

[PainterGuy]

Thank you very much!

Since $dy = g'(x)\,dx$, you get $f_Y(y) = f_X(x)/g'(x)$.

In that case, we need to sum the probabilities to get
$$f_Y(y) = \sum_{g(x_i)=y} \frac{f_X(x_i)}{\lvert g'(x_i) \rvert}.$$

Isn't the last expression more general? I mean it's applicable even if g is one-to-one function or not.

 

[vela]

Isn't the last expression more general? I mean it's applicable even if g is one-to-one function or not.

Yes, that's what the last paragraph of my previous post said.