# Probability Density function vs Mass function

I am a bit confused by using Probability Mass Function and Probability Density Function.

I understand that for discrete case like Bernoulli or Binomial, we call it pmf. For continuous case like normal distribution we call it pdf.

I have encounter question involving multivariate distribution for many times and I've solved them but I just want to verify my concept.

For 2 discrete random variable $X_1$ and $X_2$ with $P$($X_1$=$x_1$,$X_2$ = $x_2$) = $f(x_1,x_2)$

if we want to find $Y$ = a$X_1$ + b$X_2$, I used to do

$P(y)$ = ∑ $P$($X_1$=$x_1$,$X_2$ = ($y$-a$x_1)/b$) = ∑$f(x_1$,($y$-a$x_1$)/$b$))

for which I know P(y) is the PMF of Y

For the continuous case if we have similar problem:

For 2 continuous random variable $X_1$ and $X_2$ with PDF = $f(x_1,x_2)$

now I want to find the distribution of $Y$ = a$X_1$ + b$X_2$

From my understanding, I just need to replace the summation with integration(Please correct me if I am wrong)

So I would obtain the PDF of y as

$f_y$($y$) = ∫ $f(x_1$,($y$-a$x_1$)/$b$)) $dx_1$

Would appreciate if anyone could tell me if I am wrong and where could I be wrong. Thanks!

## 1 Answer

$\def\R{\mathbb R}$The density of $Y$ is by definition the function $f_Y \colon \R \to \R$, such that for every $\eta$ we have $P(Y < \eta) = \int_{-\infty}^\eta f_Y(y)\, dy$ We have using Fubini's theorem (supposing $b > 0$) \begin{align*} P(Y < \eta) &= P(aX_1 + bX_2 < \eta)\\ &= \int_{\{x:ax_1 + bx_2 < \eta\}} f_{X_1,X_2}(x) \, dx\\ &= \int_{-\infty}^\infty \int_{\{x_2:ax_1 + bx_2 < \eta\}} f_{X_1,X_2}(x)\, dx_2\,dx_1\\ &= \int_{-\infty}^\infty \int_{\{x_2: x_2 < b^{-1}({\eta -ax_1})\}} f_{X_1,X_2}(x)\, dx_2\, dx_1\\ &= \int_{-\infty}^\infty\int_{-\infty}^{b^{-1}(\eta - ax_1)} f_{X_1,X_2}(x)\, dx_2\, dx_1 & \text{substitute $y = ax_1 + bx_2$}\\ &= \int_{-\infty}^\infty\int_{-\infty}^\eta f_{X_1, X_2}\bigl(x_1, b^{-1}(y - ax_1)\bigr)\; b\,dy\; dx_1\\ &= \int_{-\infty}^\eta \left(b\int_{-\infty}^\infty f_{X_1,X_2}\bigl(x_1, b^{-1}(y -ax_1)\bigr)\, dx_1 \right)\, dy \end{align*} So the density is (as you up to an factor $b$ correctly guessed) $f_Y(y) = b\int_{-\infty}^\infty f_{X_1,X_2}\bigl(x_1, b^{-1}(y -ax_1)\bigr)\, dx_1.$

Addendum: If $a \ne 0$, you can compute $f_Y$ also the other way round $f_Y(y) = a\int_{-\infty}^\infty f_{X_1,X_2}\bigl(a^{-1}(y-bx_2), x_2\bigr)\, dx_2$ If $a = b = 0$, then $Y = 0$ and so $Y$ doesn't have a density (with respect to Lebesgue measure).

• I am surprised that there is a $a$ and a $b$ in front... Where have I gone wrong with my working above? Thanks! – K_R Sep 28 '12 at 7:23
• @K_R As you can see, the $b$ resp. $a$ comes from the substitution. In the discrete case, it won't be there. Analogies can guide you, but you should always check. Without the $a$ resp. $b$ it isn't a density, as the integral isn't 1 then. – martini Sep 28 '12 at 7:44