WEBVTT
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Let us now consider an
application of what we have
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done so far.
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Let X be a normal random
variable with
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given mean and variance.
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This means that the PDF of X
takes the familiar form.
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We consider random variable Y,
which is a linear function of
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X. And to avoid trivialities,
we assume that a is
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different than zero.
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We will just use the
formula that we
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have already developed.
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So we have that the density of
Y is equal to 1 over the
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absolute value of a.
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And then we have the density of
X, but evaluated at x equal
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to this expression.
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So this expression will
go in the place
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of x in this formula.
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And we have y minus b over a
minus mu squared divided by 2
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sigma squared.
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And now we collect these
constant terms here.
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And then in the exponent, we
multiply by a squared the
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numerator and the denominator,
which gives us this form here.
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We recognize that this is
again, a normal PDF.
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It's a function of y.
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We have a random variable
Y. This is
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the mean of the normal.
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And this is the variance
of that normal.
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So the conclusion is that the
random variable Y is normal
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with mean equal to
b plus a mu.
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And with variance a squared,
sigma squared.
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The fact that this is the mean
and this is the variance of Y
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is not surprising.
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This is how means and variances
behave when you form
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linear functions.
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The interesting part is that
the random variable Y is
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actually normal.
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Intuitively, what happened here
is that we started with a
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normal bell shaped curve.
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A bell shaped PDF for X. We
scale it vertically and
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horizontally, and then shift
it horizontally by b.
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As we do these operations, the
PDF still remains bell shaped.
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And so the final PDF is again
a bell shaped normal PDF.