Question

2+2+2+2+2+2+2+2+2+2+2+2+2+2√2π2

Answer 1

Hi there

Your answer is :-

2+2+2+2+2+2+2+2+2+2+2+2+2+2√2π2

26 + 4√2π

2 ( 13 + 2√2π ) ( Ans )


Thank You

Answer 2

X has a uniform distribution on [−π/2, π/2] what is the density of
Y = tan X?
Change of variables method. Solve Y = tan(X) to get X =
arctan(Y ). Then
fY (y) = fX(arctan(y))/|dy/dx|
Now fX(x) = 1(−π/2 < x < π/2)/π and an inverse tan-
gent is always between −π/2 and π/2 so fX(arctan(y)) =
1/π. Then dy/dx = 1 + tan2
(x) has to be evaluated at
x = arctan(y) which just gives 1 + y
2
. Thus
fY (y) = 1
π(1 + y
2
)
.
CDF method.
P(Y ≤ y) = P(tan(X) ≤ y)
= P(X ≤ arctan(y))
=



0 arctan(y) ≤ −π/2
(arctan(y) + π/2)/π −π/2 < arctan(y)π/2
1 arctan(y) ≥ π/2
Again an inverse tangent is always between −π/2 and π/2
so only the middle case counts. Now differentiate arctan(y)
to get
fY (y) = 1
π(1 + y
2
)