HW Read 6.1, with as
much of the computational detail as you can stand; at least the
"intuitive" examples.
A. Write up, coherently, the derivations of W = X+Y for f(x,y) = 1 on
the unit square 0<x, y <1. (outlined below)
B. From the handout, carefully graph the final f(w) = w2, 0<w<1; f(w) = w(2-w), 1<w<2.
C. Hand in: Find f(w) for W = X+Y for
f(x,y) = 2e-x-y , 0<x<y. Use the
f(w) formula; Integrate with respect to x. The only tricky
thing is figuring out what the upper limit on the dx interval
is: it is where the boundary line y=x intersects the line
w=x+y. For a fixed w, (say w=3) eliminate y, finding the upper
limit of the x-interval as a number. Repeat for a fixed wo.
You should get a recognizable distribution for f(w).
Optional--do it the hard way by integrating to find F(w) first.
p. 207, Read all problems, guess answers, check with back of book (good
review of many distributions.)
With 90% probability, this is the last homework assignment.
= = = = = = = = = = = = = = = = = = = = =
Continuing with Two random variables X, Y "jointly distributed".
W = X + Y
We know how to find the mean and the
variance of W = X+Y.
--Last semester you were told that the sum of two
Normal random variables is normal.
--It seems reasonable that the sum of
two independent Binomial random variables X B(n,p)
and Y B(m,p), with the same p, should be X+Y Binomial
B(n+m, p), since the first models flipping a coin n times and the
second models flipping the coin m times. Doing them in sequence
should be the same as flipping the coin m+n time.
--Likewise it seems reasonable that the sum of two
independent Poisson random variables X+Y should be
Poisson. Suppose that X counts the number of arrivals
over a time length lambda x and that Y counts the
number of arrivals over a time length lambday (then
for each the expected number per unit interval is just
1). The sum X+Y will count the number of arrivals over a
time length (lambdax + lambday) . (Ash
gives a different, also good, rationale, p. 202)
--Exponential-->Gamma: Exponential is waiting time
to first Poisson "hit"; Gamma is waiting time to n'th. We used that
Gamma Y = sum of n independent exponential X's to rationalize the mean
and s.d. of Gamma.
Question: How would you find
the distribution of W = X+Y, in general?
Try it on discrete: example from joint
discrete handouts MultDisc p. 4, ConditDiscrete
p.1.
f(x,y) = (x+y)/21, x = 1,2,3, y= 1,2. (recall sum of 2 dice)
w: 2
, 3
,
4
,
5
f: (1+1), (1+2)+(2+1), (2+2)+(3+1), (3+2), all over 21
Answer: For f(w) for a fixed w, Add
on lines x+y = w, --or-- over x, with y = w-x. i.e.
For each wo, Sum
f(x, wo-x) over all the x's for which f(x,wo-x)
is positive.
(Ch. 6.1 does more of the discrete.)
Continuous: Use method of CDF,
transforming random variables, find that we can get f(w) by
integrating on
lines x+y = w, or y = w-x. Details on Handout X + Y (Ash gives a dx
argument)
Find FW(wo) =
P(W < wo) = P(X+Y < wo), area below and to
left of x + y = wo line.
To get f(w), take derivative of F. After some hassle/trickery,
fW(wo) = -ooSoo f(x, wo-x) dx. (limits of integration depend on what x's have positive density for that wo)
Notice we're not exactly finding the Area in the x + y = wo
slice; for the area we'd be integrating the height f(x,y) by a
base
interval of width dw. We're integrating dx instead. It
works....
If X and Y are independent
with densities f(x) and g(y), the resulting integral is called the Convolution
of the two marginals: Important enough for its own notation, f*g = -ooSoo f(x)g(wo-x) dx. These occur all over mathematics, especially
when dealing with linear models.
DPGraph shows (Normal and X+Y folder) slices of the areas that get integrated in this process, for our familiar examples f(x,y) = x+y on unit square, f(x,y) = 2e-x-y on the triangular region.
Try this one by geometry, then integration: Uniform
distribution on square: f(x, y) = 1, 0< x, y < 1.
To get the DPGraph for f(x,y) = 1 on the unit square from the
one for f(x,y)= x+y, use this function as the last line of the Edit
file:
graph3d((z <1, z
<1& c>x+y))
Note need a different form for 0< w < 1 and 1< w< 2:
0< w < 1: Geometry: FW(wo)
= P(W < wo) = P(X+Y < wo) = volume
above triangle with base wo2
=
(wo2 )/2. fW(wo)
= derivative of F = wo.
Integration: FW(wo)
= (integrate x from 0 to wo }{integrate y from 0 to (wo-x)}
1 dy dx
= (integrate x from 0
to wo }[wo -x] dx = [wox - (x2
)/2 ] {endpoints 0 to wo} = (wo2
)/2. Yes!
fW(wo) = -ooSoo f(x, wo-x) dx =
(integrate x from 0 to wo }1 dx = [x] {endpoints 0 to wo}
= wo. Yes!
1< w< 2: Geometry: FW(wo) =
P(W < wo) = P(X+Y < wo) = 1 - volume above top
triangle, with base (2-wo)2
=
1 - (2-wo)2 /2 = -1 + 2wo -(wo2
)/2. fW(wo) =
derivative of F = 2-wo.
Integration: FW(wo) = (use
vertical slices first)
(Rectangle with height y = 1 and
base x = 0 to (wo-1)) + (Slant-top
quadrilateral with base x = (wo-x) to 1)
Rectangle: (integrate x from 0 to (wo-1)
}{integrate y from 0 to 1} 1 dy dx
= (integrate x from 0 to (wo-1) }1 dx = [x]
{endpoints 0 to (wo-1) } = (wo-1)
Yes.
Slant-top: (integrate x from (wo-1) to 1
}{integrate y from 0 to (wo-x)} 1 dy dx
= (integrate x from (wo-1) to 1 } (wo-x)dx
= [wox - (x2 )/2 ] {endpoints (wo-1)
to 1} = (wo - ½) - [( [wo(wo-1)
- ((wo-1)2 )/2 ]
= (if you're careful) (wo - (wo2
)/2) So
FW(wo)
= (wo-1) + (wo - (wo2
)/2) = -1 + 2wo -(wo2 )/2.
Yes.
fW(wo) = -ooSoo f(x, wo-x) dx =
(integrate x from (wo-1) to 1 }1 dx = [x] {endpoints (wo-1)
to 1} = 2 -wo. Yes!
Notice that X+Y here is the sum of two independent uniform
R.V.'s; which we saw in Math 251 had this triangular
distribution. "Sum of two spinners" Data
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