Let X~N(0,1) and let Y be independent of X and let f_Y (y)=(1/2)^(m⁄2)/( (Γ)(m/2)) y^((m⁄2)-1) e^(-m/2)  ∙I_((0,∞) ) (y). Find the distribution of T=X/√(Y/m)   .

We are given that \( X \sim N(0, 1) \), and \( Y \) is independent of \( X \) with its probability density function \( f_Y(y) = \frac{1}{2^{m/2} \Gamma(m/2)} y^{(m/2)-1} e^{-y/2} \cdot \mathbb{I}_{(0, \infty)}(y) \), which is the density of a Gamma-distributed random variable with shape parameter \( \frac{m}{2} \) and rate parameter \( \frac{1}{2} \), i.e., \( Y \sim \text{Gamma}\left(\frac{m}{2}, \frac{1}{2}\right) \).

We are tasked with finding the distribution of the random variable \( T = \frac{X}{\sqrt{Y/m}} \).

### Step-by-step Solution:

1. **Distribution of \( X \):**
   Since \( X \sim N(0,1) \), it is a standard normal random variable with the density function:
   \[
   f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}.
   \]

2. **Distribution of \( Y \):**
   The random variable \( Y \sim \text{Gamma}\left(\frac{m}{2}, \frac{1}{2}\right) \), which has the density:
   \[
   f_Y(y) = \frac{1}{2^{m/2} \Gamma(m/2)} y^{(m/2) - 1} e^{-y/2}, \quad y > 0.
   \]

3. **Transformation to \( T \):**
   We define the transformation \( T = \frac{X}{\sqrt{Y/m}} \). To find the distribution of \( T \), observe that:
   \[
   T = \frac{X}{\sqrt{Y/m}} \implies T^2 = \frac{X^2}{Y/m}.
   \]
   This suggests that \( T \) is related to the ratio of a normal variable to the square root of a Gamma variable, which is a well-known form in statistics.

4. **Chi-Squared and F-Distribution Connection:**
   - The square of a standard normal variable, \( X^2 \), follows a chi-squared distribution with 1 degree of freedom, i.e., \( X^2 \sim \chi^2_1 \).
   - A Gamma distribution \( Y \sim \Gamma(m/2, 1/2) \) is equivalent to \( Y \sim \frac{1}{2} \cdot \chi^2_m \), meaning that \( Y/m \sim \frac{1}{m} \chi^2_m \).

   Therefore, the statistic \( \frac{T^2}{1} = \frac{X^2}{Y/m} \) follows an **F-distribution** with 1 and \( m \) degrees of freedom:
   \[
   T^2 \sim F(1, m).
   \]

5. **Distribution of \( T \):**
   Since \( T^2 \sim F(1, m) \), we conclude that \( T \) follows a **t-distribution** with \( m \) degrees of freedom:
   \[
   T \sim t_m.
   \]

### Final Answer:
The random variable \( T = \frac{X}{\sqrt{Y/m}} \) follows a **Student's t-distribution** with \( m \) degrees of freedom, i.e., \( T \sim t_m \).

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Would you like any further details on the steps or derivations?

### 5 Related Questions:
1. How does the F-distribution relate to the t-distribution?
2. What is the significance of the gamma distribution in this context?
3. Can the chi-squared distribution be derived from the normal distribution?
4. How do we transform variables to find new distributions?
5. What are other applications of the t-distribution in statistics?

### Tip:
The t-distribution becomes more similar to the normal distribution as the degrees of freedom \( m \) increase. For large \( m \), \( t_m \) approximates \( N(0,1) \).

Let X~N(0,1) and let Y be independent of X and let f_Y (y) = (1/2)^(m⁄2)/(Γ(m/2)) y^((m⁄2)-1) e^(-y/2) ∙ I_(0,∞)(y). Find the distribution of T=X/√(Y/m).

Explore the detailed solution of finding the distribution of T = X/√(Y/m) where X~N(0,1) and Y is a Gamma-distributed variable. This solution covers key probability concepts, including normal and gamma distributions, transformations of variables, and concludes that T follows a t-distribution with m degrees of freedom.

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