Cheenta
Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

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?

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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

1. Arnab Chattopadhyay. says:

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.

1. Srijit Mukherjee says:

Great Work Arnab. Stay Tuned!

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy