Stochastic_Jimmy
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- Jan 10, 2013
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Hi all. Here's a problem I'm working on and I would really appreciate any tips or confirmation that I'm on the right track.
In a certain class, suppose that each student has a "true ability"(as judged by the instructor) to answer questions such that the distribution of the true ability in the class is approximately normal with mean \(\displaystyle \mu \) and variance \(\displaystyle \tau^2\). Also, suppose that the quality of the answer that a student can provide varies from lecture to lecture, such that on a random day that the class meets the quality of his/her answer is a normal r.v. with mean equal to his/her true ability and variance \(\displaystyle \sigma^2 \). The instructor comes to class on a certain day and randomly chooses a student to answer a question. Denote the true ability of the student by \(\displaystyle T\) and the quality of the answer of the student by \(\displaystyle X\). The instructor would have liked to know \(\displaystyle T\), but he can only see \(\displaystyle X\) -- and try to make inference about the unobserved true ability of the student.
Find the following:
(a) \(\displaystyle E(X \left.\right| T = t) \) and \(\displaystyle Var(X \left.\right| T = t) \)
(b) \(\displaystyle E(X) \) and \(\displaystyle Var(X) \)
(c) \(\displaystyle f_X(x) \), the marginal PDF of \(\displaystyle X\)
(d) The conditional PDF of \(\displaystyle T\) given \(\displaystyle X = x \).
Here's my attempt:
(a) We're told that the quality of a student's answer is approx normal with mean equal to his/her true ability, so then \(\displaystyle E(X\left.\right|T = t) = t \). And \(\displaystyle Var(X|T = t) = \sigma^2 \)
(b) \(\displaystyle E(X) = E[E(X|T)] = \int_{-\infty}^{\infty} E(X|T = t)\cdot f_T(t)\, dt = \int_{-\infty}^{\infty} t \cdot f_T(t) \, dt = E(T) = \mu \).
And \(\displaystyle Var(X) = E[Var(X|T)] + Var[E(X|T)] = \int_{-\infty}^\infty \sigma^2 \, f_T(t) \,dt + \int_{-\infty}^\infty (t-\mu)^2 \, f_T(t) \, dt = \sigma^2 + \tau^2 \).
(c) Since we now have \(\displaystyle E(X) = \mu, \: Var(X) = \sigma^2 + \tau^2 \) and we know that \(\displaystyle X \) is normal, we have \(\displaystyle f_X(x) = \frac{1}{\sqrt{2\pi}\sqrt{\sigma^2 + \tau^2} } \exp{\left(-\frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)}\right)} \).
(d) We have:
\(\displaystyle \begin{align*} f_{T|X}(t|x) &= \dfrac{f_{X|T}(x|t) \cdot f_T(t)}{f_X(x)} \\[5pt]
&= \dfrac{ \frac{1}{\sigma \sqrt{2\pi}}\exp{\left(-\frac{1}{2}\left(\frac{x - t}{\sigma} \right)^2\right)} \cdot \frac{1}{\tau \sqrt{2\pi}}\exp{\left(-\frac{1}{2}\left(\frac{t - \mu}{\tau} \right)^2 \right)}
}{\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2 + \tau^2} } \exp{\left(-\frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)}\right)}} \\[5pt]
&= \frac{\sqrt{\sigma^2 + \tau^2}}{\sigma \tau \sqrt{2\pi}} \exp{\left( \frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)} - \frac{(t-\mu)^2}{2\tau^2} - \frac{(x-t)^2}{2\sigma^2} \right)} \end{align*}\)
Am I approaching this problem correctly? Thank you so much for any advice/comments. I really appreciate it!
In a certain class, suppose that each student has a "true ability"(as judged by the instructor) to answer questions such that the distribution of the true ability in the class is approximately normal with mean \(\displaystyle \mu \) and variance \(\displaystyle \tau^2\). Also, suppose that the quality of the answer that a student can provide varies from lecture to lecture, such that on a random day that the class meets the quality of his/her answer is a normal r.v. with mean equal to his/her true ability and variance \(\displaystyle \sigma^2 \). The instructor comes to class on a certain day and randomly chooses a student to answer a question. Denote the true ability of the student by \(\displaystyle T\) and the quality of the answer of the student by \(\displaystyle X\). The instructor would have liked to know \(\displaystyle T\), but he can only see \(\displaystyle X\) -- and try to make inference about the unobserved true ability of the student.
Find the following:
(a) \(\displaystyle E(X \left.\right| T = t) \) and \(\displaystyle Var(X \left.\right| T = t) \)
(b) \(\displaystyle E(X) \) and \(\displaystyle Var(X) \)
(c) \(\displaystyle f_X(x) \), the marginal PDF of \(\displaystyle X\)
(d) The conditional PDF of \(\displaystyle T\) given \(\displaystyle X = x \).
Here's my attempt:
(a) We're told that the quality of a student's answer is approx normal with mean equal to his/her true ability, so then \(\displaystyle E(X\left.\right|T = t) = t \). And \(\displaystyle Var(X|T = t) = \sigma^2 \)
(b) \(\displaystyle E(X) = E[E(X|T)] = \int_{-\infty}^{\infty} E(X|T = t)\cdot f_T(t)\, dt = \int_{-\infty}^{\infty} t \cdot f_T(t) \, dt = E(T) = \mu \).
And \(\displaystyle Var(X) = E[Var(X|T)] + Var[E(X|T)] = \int_{-\infty}^\infty \sigma^2 \, f_T(t) \,dt + \int_{-\infty}^\infty (t-\mu)^2 \, f_T(t) \, dt = \sigma^2 + \tau^2 \).
(c) Since we now have \(\displaystyle E(X) = \mu, \: Var(X) = \sigma^2 + \tau^2 \) and we know that \(\displaystyle X \) is normal, we have \(\displaystyle f_X(x) = \frac{1}{\sqrt{2\pi}\sqrt{\sigma^2 + \tau^2} } \exp{\left(-\frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)}\right)} \).
(d) We have:
\(\displaystyle \begin{align*} f_{T|X}(t|x) &= \dfrac{f_{X|T}(x|t) \cdot f_T(t)}{f_X(x)} \\[5pt]
&= \dfrac{ \frac{1}{\sigma \sqrt{2\pi}}\exp{\left(-\frac{1}{2}\left(\frac{x - t}{\sigma} \right)^2\right)} \cdot \frac{1}{\tau \sqrt{2\pi}}\exp{\left(-\frac{1}{2}\left(\frac{t - \mu}{\tau} \right)^2 \right)}
}{\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2 + \tau^2} } \exp{\left(-\frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)}\right)}} \\[5pt]
&= \frac{\sqrt{\sigma^2 + \tau^2}}{\sigma \tau \sqrt{2\pi}} \exp{\left( \frac{(x-\mu)^2}{2(\sigma^2 + \tau^2)} - \frac{(t-\mu)^2}{2\tau^2} - \frac{(x-t)^2}{2\sigma^2} \right)} \end{align*}\)
Am I approaching this problem correctly? Thank you so much for any advice/comments. I really appreciate it!
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