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Size, Power, and Condition | ISI MStat 2019 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination, 2019. This primarily tests one's familiarity with size, power of a test and whether he/she is able to condition an event properly.

The Problem:

Let Z be a random variable with probability density function

f(z)=\frac{1}{2} e^{-|z- \mu|} , z \in \mathbb{R} with parameter \mu \in \mathbb{R}. Suppose, we observe X = max (0,Z).

(a)Find the constant c such that the test that "rejects when X>c" has size 0.05 for the null hypothesis H_0 : \mu=0.

(b)Find the power of this test against the alternative hypothesis H_1: \mu =2.

Prerequisites:

  • A thorough knowledge about the size and power of a test
  • Having a good sense of conditioning whenever a function (like max()) is defined piecewise.

And believe me as Joe Blitzstein says: "Conditioning is the soul of statistics"

Solution:

(a) If you know what size of a test means, then you can easily write down the condition mentioned in part(a) in mathematical terms.

It simply means P_{H_0}(X>c)=0.05

Now, under H_0, \mu=0.

So, we have the pdf of Z as f(z)=\frac{1}{2} e^{-|z|}

As the support of Z is \mathbb{R}, we can partition it in \{Z \ge 0,Z <0 \}.

Now, let's condition based on this partition. So, we have:

P_{H_0}(X > c)=P_{H_0}(X>c , Z \ge 0)+ P_{H_0}(X>c, Z<0) =P_{H_0}(X>c , Z \ge 0) =P_{H_0}(Z > c)

Do, you understand the last equality? (Try to convince yourself why)

So, P_{H_0}(X >c)=P_{H_0}(Z > c)=\int_{c}^{\infty} \frac{1}{2} e^{-|z|} dz = \frac{1}{2}e^{-c}

Equating \frac{1}{2}e^{-c} with 0.05, we get c= \ln{10}

(b) The second part is just mere calculation given already you know the value of c.

Power of test against H_1 is given by:

P_{H_1}(X>\ln{10})=P_{H_1}(Z > \ln{10})=\int_{\ln{10}}^{\infty} \frac{1}{2} e^{-|z-2|} dz = \frac{e^2}{20}

Try out this one:

The pdf occurring in this problem is an example of a Laplace distribution.Look it up on the internet if you are not aware and go through its properties.

Suppose you have a random variable V which follows Exponential Distribution with mean 1.

Let I be a Bernoulli(\frac{1}{2}) random variable. It is given that I,V are independent.

Can you find a function h (which is also a random variable), h=h(I,V) ( a continuous function of I and V) such that h has the standard Laplace distribution?

This is a problem from the ISI MStat Entrance Examination, 2019. This primarily tests one's familiarity with size, power of a test and whether he/she is able to condition an event properly.

The Problem:

Let Z be a random variable with probability density function

f(z)=\frac{1}{2} e^{-|z- \mu|} , z \in \mathbb{R} with parameter \mu \in \mathbb{R}. Suppose, we observe X = max (0,Z).

(a)Find the constant c such that the test that "rejects when X>c" has size 0.05 for the null hypothesis H_0 : \mu=0.

(b)Find the power of this test against the alternative hypothesis H_1: \mu =2.

Prerequisites:

  • A thorough knowledge about the size and power of a test
  • Having a good sense of conditioning whenever a function (like max()) is defined piecewise.

And believe me as Joe Blitzstein says: "Conditioning is the soul of statistics"

Solution:

(a) If you know what size of a test means, then you can easily write down the condition mentioned in part(a) in mathematical terms.

It simply means P_{H_0}(X>c)=0.05

Now, under H_0, \mu=0.

So, we have the pdf of Z as f(z)=\frac{1}{2} e^{-|z|}

As the support of Z is \mathbb{R}, we can partition it in \{Z \ge 0,Z <0 \}.

Now, let's condition based on this partition. So, we have:

P_{H_0}(X > c)=P_{H_0}(X>c , Z \ge 0)+ P_{H_0}(X>c, Z<0) =P_{H_0}(X>c , Z \ge 0) =P_{H_0}(Z > c)

Do, you understand the last equality? (Try to convince yourself why)

So, P_{H_0}(X >c)=P_{H_0}(Z > c)=\int_{c}^{\infty} \frac{1}{2} e^{-|z|} dz = \frac{1}{2}e^{-c}

Equating \frac{1}{2}e^{-c} with 0.05, we get c= \ln{10}

(b) The second part is just mere calculation given already you know the value of c.

Power of test against H_1 is given by:

P_{H_1}(X>\ln{10})=P_{H_1}(Z > \ln{10})=\int_{\ln{10}}^{\infty} \frac{1}{2} e^{-|z-2|} dz = \frac{e^2}{20}

Try out this one:

The pdf occurring in this problem is an example of a Laplace distribution.Look it up on the internet if you are not aware and go through its properties.

Suppose you have a random variable V which follows Exponential Distribution with mean 1.

Let I be a Bernoulli(\frac{1}{2}) random variable. It is given that I,V are independent.

Can you find a function h (which is also a random variable), h=h(I,V) ( a continuous function of I and V) such that h has the standard Laplace distribution?

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2 comments on “Size, Power, and Condition | ISI MStat 2019 PSB Problem 9”

  1. Take h = (2I-1) V. Now since I can take 0 or 1 and V \geq 0, so h can take any real numbers as it's value. Let x \geq 0. Then by using independence of the random variables I and V we get

        \begin{align*} \Bbb P (h \leq x) & = \Bbb P(I = 0) + \Bbb P(I=1, V \leq x) \\ & = \frac 1 2 + \Bbb P(I = 1) \Bbb P(V \leq x) \\ & = \frac 1 2 + \frac 1 2 (1 - e^{-x}) \\ & = 1 - \frac 1 2 e^{-x} \end{align*}

    Now if x < 0 then again by using independence of the random variables I and V we get

        \begin{align*} \Bbb P(h \leq x) & = \Bbb P (I = 0, V \geq -x) \\ & = \Bbb P (I=0) \Bbb P(V \geq -x) \\ & = \frac 1 2 (1 - \Bbb P(V \leq -x)) \\ & = \frac 1 2 (1 - (1-e^x)) \\ & = \frac 1 2 e^x \end{align*}

    Thus the random variable h follows standard laplace distribution.

    Reply

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