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For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

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.

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 \(H_1: \mu =2 \).

- 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"**

(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:

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

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?

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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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.

Great Work Arnab. Stay Tuned!