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I have an interesting problem, i have seen in many text books ways of calculating conditional pdfs but not many where given a set of conditional pdfs for a variable we wish to calculate it's pdf. In particular i'm interested in problems with mixed types. As an example suppose the random variable $X$ takes values in the set $\{-2,2\}$ with equal probability. Given $X=x$ the random variable $Y \sim N(x,1)$. How can i generate the pdf for $Y$? My approach would be this.

To calculate the cdf $F_{Y}(y) = P(Y \leq y) = P(Y \leq y \cap X=-2) + P(Y \leq y \cap X=2) = P(Y \leq y|X=-2)P(X=-2)+P(Y \leq y|X=2)P(X=2)$.

Since i have the conditional cdf's i can calculate $F_{Y}(y)$. Once i have $F_{Y}(y)$ i can differentiate to calculate $f_{y}(y)$. Now i know there would be no analytical solution given that the conditional distribution of $Y$ is normal but is this approach the right way to go? Are there any other techniques commonly used?

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  • $\begingroup$ I presented a solution for something like this some time back, only it was for a conditional distribution, but I think you'll find it helpful here as it applies: stats.stackexchange.com/questions/404102/… $\endgroup$ – StatsStudent May 14 at 16:24
  • $\begingroup$ Thanks, which part exactly relates to this problem? Also is my solution a correct way of approaching it? $\endgroup$ – Iltl May 15 at 8:40
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    $\begingroup$ $(X,Y)$ does not have a pdf. $Y$ does have a pdf and it's simple to write down in terms of elementary functions, leaving one to wonder what you might mean by "no analytical solution." $\endgroup$ – whuber May 22 at 18:22
  • $\begingroup$ You are absolutely on the right track. $P(Y \leq y \cap X=\pm 2)$ is exactly $\Phi(y\pm 2)$ where $\Phi(\cdot)$ is the standard normal CDF (whose derivative is $\phi(\cdot)$, the standard normal pdf). This gives $$f_Y(y) =\left.\left. \frac 12\right[\phi(y-2)+\phi(y+2)\right]$$ which is the answer obtained by StatsStudent, but with a lot less effort and use of mistaken ideas such as the pdf of $(X,Y)$. As whuber's comment points out, $(X,Y)$ does not have a (joint) pdf. $\endgroup$ – Dilip Sarwate May 23 at 3:16
  • $\begingroup$ @DilipSarwate what ideas do you see as "mistaken?" Genuinely hoping to learn. I'm also not sure this approach is any simpler -- you've just left out the mathematics, although, I could see how some might see this as more intuitive. By the way, I think your comment is good enough for an "answer." $\endgroup$ – StatsStudent May 23 at 14:23
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I think you are making this too complicated. You simply need to find:

\begin{eqnarray*} f_{Y}(y) & = & \int_{-\infty}^{\infty}f_{Y|X}(y|x)f_{X}(x)dx\\ & = & \int_{-\infty}^{\infty}f_{X,Y}(x,y)dx \end{eqnarray*}

if your goal is to obtain the PDF of $f_Y(y)$.

So in your case, you can start with:

\begin{eqnarray*} f_{X}(x) & = & \begin{cases} 1/2 & x=-2\\ 1/2 & x=2\\ 0 & otherwise \end{cases} \end{eqnarray*}

and

\begin{eqnarray*} f_{Y|X}(y|x) & = & \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-x)^{2}},\,y\in(-\infty,\infty);x={\{-2,2}\} \end{eqnarray*}

So, you can compute:

\begin{eqnarray*} f_{X,Y}(x,y) & = & f_{Y|X}(y|x)f_{X}(x)\\ & = & \begin{cases} \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}(y+2)^{2}} & x=-2,\,y\in(-\infty,\infty)\\ \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}(y-2)^{2}} & x=2,\,y\in(-\infty,\infty)\\ 0 & \text{otherwise} \end{cases} \end{eqnarray*}

Now, you can find the marginal PDF, $f_Y(y)$ over $y\in(-\infty,\infty)$ by "summing out" $X$: \begin{eqnarray*} f_{Y}(y) & = & \sum_{x=\{-1,2\}}f_{X,Y}(x,y)\\ & = & f_{X,Y}(-2,y)+f_{X,Y}(2,y)\\ & = & \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}(y+2)^{2}}+\frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}(y-2)^{2}}\\ & = & \frac{1}{2}\left[\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}[(y+2)-0]^{2}}+\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}[(y-2)-0]^{2}}\right]\\ & = & \frac{1}{2}\left[\phi(y+2)+\phi(y-2)\right] \end{eqnarray*}

where $\phi(\cdot)$ is the PDF of a Standard Normal Random Variable.

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    $\begingroup$ can you actually mix pmfs and pdfs like this? What you describe as $f_{X}(x)$ isn't actually a probability density function. $\endgroup$ – Iltl May 22 at 15:26
  • $\begingroup$ $f_X(x)$ here is a PMF. $f_{Y|X}(y|x)$ is a PDF. I think you'd benefit from a review of Section 8-3 of Carol Ash's "Probability Tutoring book." Specifically review Example 2: "First stage discrete, second stage continuous" and problem 3 in that section: bit.ly/30uEyYm. $\endgroup$ – StatsStudent May 22 at 17:10
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    $\begingroup$ Errrrr No. Your mistake lies in the third line of the $f_Y(y)$ computation where you set $\exp(a) + \exp(b)$ to $\exp(a+b)$. The unconditional pdf of $Y$ is the two-humped camel $\left.\left. \frac 12 \right[\phi(y-2) + \phi(y+2)\right]$ (where $\phi(\cdot)$ is the standard normal pdf) and not what you have written. Put another way, $Y$ has a mixture pdf where the mixture is of two normal pdfs $N(2,1)$ and $N(-2,1)$ with equal weights. $\endgroup$ – Dilip Sarwate May 22 at 18:12
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    $\begingroup$ haha, ugh. Yes, I totally messed up some very, very simple algebra (I blame my terrible handwriting ;-) ). I'll fix when back at my desk. Thanks for keeping me straight. The rest of the work previous to this step should still stand. Thanks, @DilipSarwate. I could have been more precise with my language: a mixture pdf indeed is the preferred term. $\endgroup$ – StatsStudent May 22 at 18:32
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    $\begingroup$ Thanks for the link to the great book, do you have a copy with no missing pages? $\endgroup$ – Iltl May 23 at 8:45

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