Sum of weighted chi square distributions Let $X_1 \sim \chi_{k}^2$ and $X_2 \sim \chi_{k}^2$ be i.i.d and both $a_1$ and $a_2$ positive real values.
How can be expressed the PDF of $Y = a_1X_1 + a_2X_2$? Is it also a chi-square distribution?
thanks

A linear combination of chi-squared random variables is not, in general, chi-squared
The answer is NO unless the $a_i$ are either 0 or 1. This is easy to see by looking at moment generating functions. (Also, by looking
at means and variances, see Addendum.)
Such distributions occur often in practice.
The issue of the distribution $Y$ is of interest in many
practical situations. An elementary example is in trying to 
find a confidence interval (CI) for the 'batch' variance $\sigma_B$ in a
random-effects one-factor ANOVA model $Y_{ij} = \mu + B_{i} + e_{ij},$
where $i = 1,\dots,b$ batches, $j = 1, \dots, r$ replications
per batch, $B_i \text{ iid Norm(}0, \sigma_B^2\text{)}$ and
$e_{ij} \text{i id Norm (}0, \sigma_e^2\text{)}.$ 
The error mean
square in the ANOVA table is an unbiased estimate of $\sigma_e^2$ and a CI for it can be
constructed using an obvious chi-squared distribution. However,
the mean square for batches is influenced by both $\sigma_A^2$
and $\sigma_e^2.$ 
A method of moments estimate of $\sigma_A^2$
is a linear combination of the two mean squared terms and its distribution
would be a linear combination of chi-squared distributions. 
You might look at some of the methods that have
been attempted to approximate the distribution and find CIs for
batch variances. In this case, one of the $a_i$s is negative. (One very nice solution to the problem involves a
Bayesian formulation of the ANOVA and using a Gibbs sampler
to get an interval estimate, but that has nothing to do with
solving your problem.)
Addendum with simulation: Below is a brief simulation of 100,000 realizations of $S = X_1 + X_2,$
where $X_1 \sim \text{Chisq}(5), X_2 \sim \text{Chisq}(10)$ and
also of $T = 3X_1 + 0.5X_2.$ The sample of $S$ passes a Kolmogrorv-
Smirnov goodness-of-fit test for matching $\text{Chisq}(15)$ and
also has mean and variance compatible with this distribution.
However, there can be no chi-squared distribution with mean
$E(T) = 3(5) + 0.5(10) = 20$ and variance $V(T) = 9(10) + 0.25(20) = 95 \ne 40.$
In practical applications, distributions of linear combinations
of chi-squared random variables are often approximated by simulation.
 x1 = rchisq(10^5, 5);  x2 = rchisq(10^5, 10)
 s = x1 + x2;  mean(s);  var(s)
 ## 15.02936  # E(S) = 15 as for Chisq(15)
 ## 30.07302  # V(S) = 30 as for Chisq(15)
 t = 3*x1 + .5*x2;  mean(t);  var(t)
 ## 20.00794  # Not a possible combination of
 ## 94.5343   #   mean and var for a chisq distn.

