Probability density function homework

[kingwinner]

Find a constant c such that f(x,y)=cx² + e^-y, -1<x<1, y>0, is a proper probability density function.

My idea:
f(y)
1
=∫ f(x,y) dx
-1

So I have found f(y), now I set the following integral equal to 1 in order to solve for c:

∞
∫ f(y) dy = 1
0

Integrating, I get something like (c)(∞)+...=1

If this is the case, how can I solve for c? (can't divide something by infinity) Is this question even possible?

Thanks!

 

[Dale]

Try c=0. It is the only number that could possibly have a finite limit when multiplied by infinity.

 

[EnumaElish]

1
∫ f(x) dx = 1
-1

Integrating, I get something like (c)(∞)+...=1

How? Can you describe?

 

[kingwinner]

How? Can you describe?

Performing the integartion, I get
f(y)=
1
∫ cx² + e^-y dx
-1
=2c/3 + 2e^-y

Setting
∞
∫ f(y)dy=1
0
I get 2c/3 (∞) + 2 =1

 

[Dale]

=2c/3 + 2e^-y

Look at your 2c/3 term, it is a constant wrt y. Any constant, integrated from 0 to infinity, is infinite. It must be zero in order for the integral to converge.

 

[kingwinner]

Look at your 2c/3 term, it is a constant wrt y. Any constant, integrated from 0 to infinity, is infinite. It must be zero in order for the integral to converge.

For 2c/3 (∞) + 2 =1
Put c=0
=> 2=1 (if the first term is zero)
So c=0 doesn't give a proper probability density function...

 

[EnumaElish]

Find a constant c such that f(x,y)=cx² + e^-y, -1<x<1, y>0, is a proper probability density function.

No such real c exists.