Marginal pdf, what am I doing wrong?

Rifscape
Messages
41
Reaction score
0

Homework Statement



f(xy)=49/8*e^(−3.5*y) 0 < y < inf and −y < x < y

0 otherwise

a. Find the marginal probability density function of X, fX(x). Enter a formula in the first box, and a number for the second and the third box corresponding to the range of x. Use * for multiplication, / for division, ^ for power, abs for absolute value and exp for exponential function. For example, 3abs(x-5)exp(-x/2) means 3|x-5|e-x/2. Use inf for ∞ and -inf for -∞.

I'm not sure what I am doing wrong here, I keep getting 7/4 and the bounds are from -inf to inf right?

Homework Equations


The marginal pdf equation for x.

The Attempt at a Solution



I did the 8/49*∫e^(-3.5*y) from 0 to inf and got 7/4 with the bounds of x being from 0 to inf. What exactly am I doing wrong?

Or would the bounds be from x to infinity? Any advice is appreciated.

Thanks for the help
 
Last edited:
Physics news on Phys.org
First of all I think you should check the problem is typed correctly. I don't think what you have given is a density function. Second, draw a picture of the region where ##f(x,y)\ne 0##. You need to know that to know the correct limits for your integrals.
 
LCKurtz said:
First of all I think you should check the problem is typed correctly. I don't think what you have given is a density function. Second, draw a picture of the region where ##f(x,y)\ne 0##. You need to know that to know the correct limits for your integrals.

You're right, it was 49/8 not 8/49 for the constant. I tried drawing the picture, I got that the bounds seem to look like its from -inf to inf. But for the distribution wouldn't I have to get x in terms of y? That's why I thought the bounds would be from -x to x, since x < y and y > -x.
 
Rifscape said:
You're right, it was 49/8 not 8/49 for the constant. I tried drawing the picture, I got that the bounds seem to look like its from -inf to inf. But for the distribution wouldn't I have to get x in terms of y? That's why I thought the bounds would be from -x to x, since x < y and y > -x.
Your original statement of the problem had limits: 0 < y < inf and −y < x < y. That is the region where your ##f(x,y)\ne 0##. Now, the definition of the marginal density is$$
f_X(x) = \int_{-\infty}^\infty f(x,y)~dy$$You only have to integrate over the region where ##f(x,y)\ne 0##. For each ##x## you integrate in the ##y## direction. When ##x<0## where does ##y## go? When ##x>0## where does ##y## go? You need to look at your region. So let's see how you set up that integral.
 
LCKurtz said:
Your original statement of the problem had limits: 0 < y < inf and −y < x < y. That is the region where your ##f(x,y)\ne 0##. Now, the definition of the marginal density is$$
f_X(x) = \int_{-\infty}^\infty f(x,y)~dy$$You only have to integrate over the region where ##f(x,y)\ne 0##. For each ##x## you integrate in the ##y## direction. When ##x<0## where does ##y## go? When ##x>0## where does ##y## go? You need to look at your region. So let's see how you set up that integral.

Would I need to split it up and integrate x from -inf to 0, and then integrate x from 0 to inf?
 
LCKurtz said:
When ##x<0## where does ##y## go? When ##x>0## where does ##y## go? You need to look at your region. So let's see how you set up that integral.

Rifscape said:
Would I need to split it up and integrate x from -inf to 0, and then integrate x from 0 to inf?

You aren't integrating ##x##. It it is a ##y## integral. You could start by answering the two questions I asked you above. It would also help if you would describe the region in words so I know you have the correct region in the first place.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top