sm10_22

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Problem 10.22. Show that the choice / 2γ µ= minimizes the probability of error given 
by Eq. (10.26). Hint: The Q-function is continuously differentiable. 
 
Solution
From (10.26), we have the average probability of error as: 
 
⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛ −= σ
γ
σ
γµγ QQPe 2
1
2
1)( 
 
Recall the definition of Q-function: 
∫
∫
−
∞−
+∞
−=
−=
−=
x
x
duu
su
dssxQ
)2/exp(
2
1
)let(
)2/exp(
2
1)(
2
2
π
π
 
 
So the derivative is given by 
 
0)2/exp(
2
1)( 2 ≤−−= x
dx
xdQ
π 
 
Substituting this result into the definition of Pe(γ) we obtain 
 
2 2
2 2
2 2
2 2
( ) 1 1 ( ) 1 1 1 1exp exp
2 2 2 22 2
1 ( )exp exp
2 22 2
edP
d
γ µ γ γ
γ σ σ σπ π
µ γ γ
σ σπσ
⎛ ⎞ ⎛− − − −= ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅⎜ ⎟ ⎜⎝ ⎠ ⎝
⎧ ⎫⎛ ⎞ ⎛ ⎞−⎪ ⎪= − − −⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭
σ
⎞⎟⎠
 
Setting 0
)( =γ
γ
d
dPe implies 
 
2 2
2 2
2 2
( )exp exp
2 2
( )
/ 2
µ γ γ
σ σ
µ γ γ
γ µ
⎛ ⎞ ⎛−− = −⎜ ⎟ ⎜⎝ ⎠ ⎝
− =
=
⎞⎟⎠
 
 
 
 
Continued on next slide 
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Problem 10.22 continued 
 
Checking the second derivative, we have 
 
2 2 2
2 2 2 2
( ) 1 2( ) ( ) 2exp exp
2 2 22 2
0
ed P
d
γ µ γ µ γ γ γ
γ σ σ σπσ
⎡ ⎤⎛ ⎞ ⎛− −= ⋅ − + −⎢ ⎥⎜ ⎟ ⎜⎝ ⎠ ⎝⎣ ⎦
>
22σ
⎞⎟⎠ 
when γ = µ/2. Therefore at / 2γ µ= , )(γeP has a minimum value. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.