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Problem 1 2 Solution

Y1^2 has a chi-squared distribution with one degree of freedom. YT Y does not have an exact chi-squared distribution due to scaling, but instead follows a generalized chi-squared distribution. For a multivariate normal Y ~ MVN(mu, V), YTV-1Y follows a generalized chi-squared distribution whose exact form depends on mu and V.

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32 views2 pages

Problem 1 2 Solution

Y1^2 has a chi-squared distribution with one degree of freedom. YT Y does not have an exact chi-squared distribution due to scaling, but instead follows a generalized chi-squared distribution. For a multivariate normal Y ~ MVN(mu, V), YTV-1Y follows a generalized chi-squared distribution whose exact form depends on mu and V.

Uploaded by

farrukhbhatti78
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Problem 1.

2 Solution

a. What is the distribution of Y1^2?

Y1 is a normally distributed random variable with mean mu = 0 and variance sigma^2 = 1, so Y1 ~

N(0,1). The square of a standard normal random variable follows a chi-squared distribution with one

degree of freedom, denoted as chi^2(1).

b. Obtain an expression for Y^T Y. What is its distribution?

Here, Y is a vector defined as follows:

Y = [ Y1 ]

[ (Y2 - 3)/2 ]

To find Y^T Y, we take the transpose of Y and multiply it by Y:

Y^T Y = [ Y1 (Y2 - 3)/2 ] [ Y1 ]

[ ][ ]

[ ] [ (Y2 - 3)/2 ]

Y^T Y = Y1^2 + ((Y2 - 3)/2)^2

Now, Y1^2 has a chi-squared distribution with one degree of freedom as established in part (a).

The term ((Y2 - 3)/2)^2 can be rewritten as 1/4(Y2 - 3)^2. Since Y2 is normally distributed with mean

mu = 3 and variance sigma^2 = 4, the term (Y2 - 3)^2 is a scaled chi-squared distribution with one

degree of freedom, as it is the square of a standard normal random variable (after standardizing Y2).

The sum of independent chi-squared random variables is also chi-squared, with degrees of

freedom equal to the sum of their degrees of freedom. However, because the second term is scaled

by a factor of 1/4, Y^T Y does not follow a chi-squared distribution exactly due to this scaling.

Instead, it follows a generalized chi-squared distribution.

c. If Y = [ Y1 ] and its distribution is Y ~ MVN(mu, V), obtain an expression for Y^T V^-1 Y. What is
its distribution?

For a multivariate normal distribution Y ~ MVN(mu, V), the quantity Y^T V^-1 Y is known as a

quadratic form. This quantity follows a generalized chi-squared distribution. The expected value of

Y^T V^-1 Y is mu^T V^-1 mu + trace(V^-1 Sigma), where Sigma is the covariance matrix of Y, and V

is the matrix used in the quadratic form (in this case, V and Sigma are the same).

The exact distribution of Y^T V^-1 Y depends on V and mu. If mu = 0, Y^T V^-1 Y is distributed as

a chi-squared distribution with degrees of freedom equal to the rank of V. However, if mu is not

equal to 0, then Y^T V^-1 Y doesn't have a standard distribution but can still be expressed in terms

of its moment-generating function (MGF) or characteristic function.

In the general case, the distribution of Y^T V^-1 Y is not one of the standard forms and is typically

referred to as a generalized chi-squared distribution, and its moments can be computed from the

eigenvalues of V^-1/2 Sigma V^-1/2, where V^-1/2 is the matrix such that V^-1/2 V^-1/2 = V^-1.

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