MTH366 20130403 week 1 Wednesday April 3, 2013 page 1
Last time…
If we know X̲̅ and fX̲̅(x) and Y̲̅ = U(X̲̅ ), sometimes we can find fY̲̅(y).
new HW posted, due Wednesday of week 2
quiz this week: read syllabus and email teacher about it: email the components of your grade based on
the syllabus. due Friday of week 1 at 5:00 PM PDT
fX̲̅(x) is the p.d.f. for continuous variable
From the p.d.f. for Y̲̅ we can compute:
∞
the mean
μX̅ ̲ =E(Y̲̅ )= ∫ yfY̲̅ (y)dy
-∞
the variance
2
2
σ2Y̲̅ = E ((Y̲̅ − μY̲̅ ) ) = E (Y̲̅ ) − μ2Y̲̅
σY̲̅ = √σ2Y̲̅
the standard deviation
∞
since E(U(x))= ∫ U(x)fX̲̅ (x)dx
-∞
Summary of what we’ve done so far.
Let Y̲̅ be a function of continuous r.v.’s X̲̅1, X̲̅2, …X̲̅n. Then we:
1. Find the region Y̲̅ = y in the (x1, x2, … xn) space.
2. Find the region Y̲̅ ≤ y.
3. Find the distribution FY̲̅(y) = G(y) = P(Y̲̅≤y) by integrating “fX̲̅(x)” over the region Y̲̅≤y.
4. Find the density function fY̲̅(y) = G(y) by differentiating FY̲̅(y) = G(y).
If we have h(x,y,z) find
∂
d
not
∂z
dz
Let’s generalize the approach.
The method of transformations (change of variables) through the distribution function approach we can
arrive at a single method of finding the p.d.f. for Y̲̅ = U(X̲̅ ) provided U(X̲̅ ) is either increasing or decreasing.
Increasing? From calculus class we say h(x) is increasing if h’(x)>0. We say h(x) is decreasing if h’(x)<0.
Invertible? From MTH111, we had:
one to one: one output for each input
onto: everything in the range comes from something in the domain
horizontal line test for one to one
vertical line test for function
If y=x 2 then ±√y=x (not technically invertible)
Suppose U(X̲̅ ) is increasing and that Y̲̅ = U(X̲̅ ) where X̲̅ has p.d.f. fX̲̅(x). If y = U(x) then U-1(y)=x. Apply U-1
to both sides so U-1(y)=U-1(U(x)) and U-1(y)=x.
So in theory FY̲̅(y) = P(Y̲̅≤y) by definition of the distribution function and also = P(U(X̲̅ )).
Aside: Start with U(x)≤y so x≤U-1(y) and FY̲̅(y) = P(X̲̅≤U-1(y)). We’re changing an unknown integration into
one we have.
FY̲̅(y)=FX̲̅ (U-1(y)) where FX̲̅ is the distribution function for X̲̅ . To find the density function (p.d.f.) of Y̲̅ we
must differentiate FY̲̅(y).
fY̲̅ (y)=
d
d
FY̲̅ (y)=
(F (U -1 (y)))
dy
dy X̲̅
d (U -1 (y))
d U(t)
d(eU ) dU
d
d
dx
dx
U(t) ' (t)=
-1 (y)))
=
(FX̲̅ (U
=
(FX̲̅ (x)) = fX̲̅ (U -1 (y))
(e )= e U
dt
dU dt dx
dy
dx
dy
dy
Note that
dx
1
=
dy
dy
( )
dx
ex)
2x 0≤x≤1
fX (x)= {
0 elsewhere
Find the p.d.f. for Y̲̅ = 3X̲̅-1
1. What is U? Is it “nice?”
U(x) = 3x-1 or y=3x-1
x=
1
(y+1)= U -1 (y)
3
dx
d 1
1
=
( y+1) =
dy dy 3
3
Then fY̲̅ (y)= fX̲̅ (U -1 (y))
1
-1
dx
= {2U (y) 3 -1≤y≤2
dy
0 elsewhere
2
= {9 (y+1) -1≤y≤2
0 elsewhere
We really need U(X̲̅ ) to be invertible if we write things like U-1(Y̲̅) so 1-1 and onto or injective and
surjective, or bijective. Then we can do the following:
FY̲̅(y) = FX̲̅(U-1(y)) and differentiating
fY̲̅ (y)=
d
d
dx
(FY̲̅ (y)) =
(FX̲̅ (x)) | |
dy
dx
dy