- #1
kingwinner
- 1,270
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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!
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!
Last edited: