# Joint PDFs and Expected Values of 3 Cont Rand Vars

1. Nov 13, 2012

### SpiffWilkie

You know how sometimes, you take a long time to write a question, and then you submit, but it takes you to a login screen and then you lose everything you wrote?

Anyhow...I would just like to verify some work here.

The problems states that there are 3 mutually independent random variables, X1, X2, X3, each with the pdf f(x) = 2x, 0 < x < 1.

a) What is the joint pdf?
For this, I'm assuming I just need to take the product of the three pdfs, giving me 8x1x2x3?

b) Find the expected value of 5X1X2X3.
Am I to assume that this is equal to 5 * E(X1) * E(X2) * E(X3)? If that is true, I would calculate 5 * (∫2x2dx)3 from 0 to 1?
This gives the result 40/27 ≈ 1.48

c) Find P(X1 < 1/2, X2 < 1/2, X3 < 1/2)
I calculated (∫2xdx)3 from 0 to 1/2, giving me 1/64 or .015625

I'm struggling trying to connect examples from class/book/internet with actual problems given, so any help is much appreciated.

Thanks,
Steve

Last edited: Nov 13, 2012
2. Nov 13, 2012

### Ray Vickson

For (b): do not *assume*. This is something you should know. The short answer is 'yes' in this case---that is, that E(X1 X2 X3) = EX1 * EX2 * EX3----- but not in general. Do you know why?

If you are unsure, the best way is to write out everything in detail. For example, you can certainly say that
$$E(X_1 X_2 X_3) = \int_0^1 \int_0^1 \int_0^1 x_1 x_2 x_3 f(x_1,x_2,x_3) \, dx_1 \, dx_2 \, dx_3,$$
and you can go on from there.

RGV

3. Nov 13, 2012

### SpiffWilkie

So, in your equation, then just integrate $$\int_0^1 2x_3^2 dx_3$$ substituting that into the integral for x2, and then that result into the integral for x1?
Actually, the explanation helped me out with another problem:

$$f(x_1) = 2x_1, 0 < x_1 < 1 and f(x_2) = 4x_2^3, 0 < x_2 < 1$$

so, $$E(X_1^2X_2^3) = \int_0^1 \int_0^1 x_1^2 x_2^3 f(x_1,x_2) \, dx_1 \, dx_2$$
$$= \int_0^1 \frac {4}{7} 2x_1^3 \,dx_1$$
$$= \frac {2}{7}$$