Question

Can someone please show me how to prove that the following integral = 1

Answer 1

\[\theta^k (-y^-k), 0<=y<=2\]

Answer 2

??
i don;t see an integral

Answer 3

??? Where is the integral ??

Answer 4

Sorry - I'll attach what I started with and my workings. I am confused about where to apply the limits (and even if the limits are correct)

Answer 5

Can you only attach the question separately because I am unable to distinguish what you have worked out and what is the question given

Answer 6

Sure no problem

Answer 7

k and theta are constants,Am I right ?

Answer 8

That is my understanding, yes

Answer 9

First of all I would like to say that you have gone wrong in your integration.

\[k \theta^{k} \int\limits_{}^{}y^{-k-1} dy= k \theta^{k}{{ y^{-k-1+1} }\over {-k+1-1}} = -\theta^{k}y^{-k}\]

Answer 10

ahh you beat me to it...and the limits are from theta to infinity

Answer 11

Sorry @SKMC you have done it correctly
@dumbcow , thanks for telling the limits.

Answer 12

@dumbcow , Can you explain how you have arrived at these limits ??

Answer 13

@shivam_bhalla ,
given theta >0 and y>theta
so min_y must be theta and max_y has no limit

Answer 14

but you have not taken into account that
for theta>0 and k>2 --> f(y)=0

Answer 15

This is the point where i became unsure how to proceed

Answer 16

oh i see, maybe i misread that part
hmm but if you assume f(y)>0 when y>theta it should still work
and the integral evaluates to 1 with those limits

Answer 17

@dumbcow, you are right. The integral evaluates to 1

Answer 18

I dont understand how that evaluates to 1

Answer 19

\[k \theta^{k}\int\limits_{\theta}^{\infty}y^{-k-1} dy = k \theta^{k} |_\theta^{\infty} \frac{y^{-k}}{-k} = k \theta^{k}(0-\frac{\theta^{-k}}{-k}) = \theta^{k-k} = 1\]

Answer 20

So from my workings, where I got \[\theta^k(-y^-k)\] my next line should have been \[\theta ^{k-k}\] instead of \[-(\theta/y)^{k}\]and from \[\theta ^{k-k}\] \[\theta =1?\]

Answer 21

anything to zero power is 1

Answer 22

Many Thanks for your help shivam_bhalla and dumbcow- i appreciate your expertise and your patience :)